Question 1

An airplane moves at 120 m/sec to the west with respect to the air.The air moves at 50 m/sec to the south with respect to the ground.How fast does the airplane move with respect to the ground?
A)70 m/s
B)120 m/s
C)130 m/s
D)170 m/s

Accepted Answer

We can use the vector addition method to solve this problem:
- Draw a right triangle with one side representing the westward velocity of the airplane (120 m/s) and another side representing the southward velocity of the air (50 m/s).
- The hypotenuse of the triangle represents the resultant velocity of the airplane with respect to the ground.
- Use Pythagorean theorem to find the magnitude of the hypotenuse: 
  magnitude = sqrt((120 m/s)^2 + (50 m/s)^2) = 130 m/s
- Use trigonometry to find the direction of the hypotenuse:
  angle = atan(50/120) = 22.6 degrees south of west
Therefore, the airplane moves at a speed of 130 m/s and at an angle of 22.6 degrees south of west with respect to the ground.

Question 2

A projectile is launched at 10.0 m/s in a direction 45.0 degrees above the horizon.It hits the ground 10.0 m below its initial position.What is its speed just before hitting the ground? Ignore air resistance.
A)14.1 m/s
B)17.2 m/s
C)20.0 m/s
D)24.1 m/s

Accepted Answer

The speed of the projectile just before hitting the ground can be found using energy conservation principles. The initial kinetic energy plus the initial potential energy equals the final kinetic energy (since we're ignoring air resistance). The initial velocity can be broken down into horizontal ($v_{x0} = 10\cos(45^\circ)$) and vertical ($v_{y0} = 10\sin(45^\circ)$) components, both of which equal $10\sqrt{2}/2$ m/s due to the 45-degree angle. The initial potential energy is $mgh$, where $h = 10$ m is the height below the launch point. The final kinetic energy must account for both the change in height and the initial kinetic energy. Solving for the final speed involves using the equation for kinetic energy ($1/2 mv^2$) and potential energy ($mgh$), and recognizing that the mass ($m$) cancels out. The calculation confirms that the speed just before hitting the ground is approximately 17.2 m/s, taking into account the gravitational acceleration ($g = 9.8$ m/s$^2$) and the initial conditions given.

Question 3

A skier leaves a ski jump with a horizontal velocity of 27.0 m/s (and no vertical velocity component).The skier lands 2.50 seconds later.What is the magnitude of the skier's displacement during the jump
A)98.2 m
B)74.1 m
C)30.7 m
D)67.5 m

Accepted Answer

Since the skier has no vertical velocity component, we can ignore the effects of gravity and assume the motion is purely horizontal. Using the equation for displacement (s = vt), we have s = (27.0 m/s)(2.50 s) = 67.5 m. However, we are looking for the magnitude of the displacement, so we take the absolute value, giving us |67.5 m| = 67.5 m. Therefore, the best choice is B, 74.1 m is incorrect because it's not the distance traveled in the horizontal direction.

Question 4

An object moves from a point with coordinates x = 0.0 m,y = -1.0 m,to a point with coordinates x = 8.0 m,y = -9.0 m in a time interval of 3.4 seconds.The magnitude of its the average velocity is
A)6.2 m/s.
B)5.0 m/s.
C)7.3 m/s.
D)4.9 m/s.
E)3.2 m/s

Accepted Answer

The answer of An object moves from a point with...

Question 5

A child kicks a football with an initial velocity of 25.0 m/s at an angle of 38S1U1P1$\circ$S1S1P0above horizontal.The ball lands at the same height from which it was kicked.What is the range of the ball?
A)24.7 m
B)39.3 m
C)51.8 m
D)61.8 m

Accepted Answer

The range of a projectile launched with an initial velocity $v_0$ at an angle $\theta$ above the horizontal, landing at the same height, is given by the formula: $$R = \frac{v_0^2 \sin(2\theta)}{g}$$ where $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$). Plugging in the given values: $v_0 = 25.0 \, \text{m/s}$, $\theta = 38^\circ$, and $g = 9.8 \, \text{m/s}^2$, we find the range $R$. Calculation shows that the correct answer is closest to 61.8 m, making option D the correct choice.

Question 6

You are walking to catch your plane and decide to take the moving sidewalk in the airport.It has an average speed of 2.0 m/s.If you are walking with a speed of 1.5 m/s and the walkway is 85.0 m long.How long will it take you to travel the 85.0 m?
A)12 s
B)24 s
C)34 s
D)36 s
E)48 s

Accepted Answer

The answer of You are walking to catch your plane...

Question 7

A skier leaves a ski jump with a horizontal velocity (and no vertical velocity component).If the magnitude of the horizontal component of her velocity the instant before she lands is 19.6 m/s,how much time elapses between leaving the jump and landing?
A)1.0 s
B)2.7 s
C)5.4 s
D)Not enough information is given to determine the time of flight.
E)2.0 s

Accepted Answer

The time of flight for a projectile depends on the vertical component of the velocity and the height from which it is launched. Since no information about the vertical component of the velocity or the height of the jump is provided, we cannot determine the time of flight.

Question 8

A skier leaves a ski jump with a horizontal velocity (and no vertical velocity component).If the magnitude of the horizontal component of her velocity the instant before she lands two seconds later is 15.0 m/s,what was the horizontal component of her initial velocity
A)24.0 m/s.
B)19.6 m/s.
C)15.0 m/s.
D)43.6 m/s.
E)0 m/s.

Accepted Answer

In the absence of horizontal forces (assuming air resistance is negligible), the horizontal component of velocity remains constant throughout the flight. Therefore, if the skier's horizontal velocity is 15.0 m/s just before landing, it was also 15.0 m/s upon leaving the ski jump.

Question 9

A skier leaves a ski jump with a horizontal velocity of 21.0 m/s (and no vertical velocity component).The skier lands three seconds later.What is the magnitude of the horizontal component of the skier's displacement during the jump
A)32.2 m
B)63.0 m
C)44.1 m
D)51.8 m

Accepted Answer

The horizontal component of the skier's displacement can be calculated using the formula $d = vt$, where $d$ is displacement, $v$ is the horizontal velocity, and $t$ is the time. Given $v = 21.0 \, \text{m/s}$ and $t = 3 \, \text{s}$, the displacement is $d = 21.0 \, \text{m/s} \times 3 \, \text{s} = 63.0 \, \text{m}$.

Question 10

If the launch speed of a kicked football was doubled,then by what factor would its hang time increase?
A)the square root of two
B)two
C)four
D)eight

Accepted Answer

The hang time of a projectile is directly proportional to its initial vertical velocity. If the launch speed is doubled, the vertical component of the velocity is also doubled, thus doubling the hang time.

Question 11

The maximum height above ground you can achieve with a projectile launched vertically upward from ground level,if you are able to give it an initial speed of 65.0 m/s,is
A)336 m.
B)215 m.
C)125 m.
D)6.36 m.

Accepted Answer

The maximum height a projectile can reach when launched vertically upward can be found using the formula:
H = (v^2)/(2g)
where v is the initial velocity and g is the acceleration due to gravity (9.81 m/s^2)

Plugging in the given values, we get:
H = (65^2)/(2*9.81) = 215 m

Therefore, the correct answer is B) 215 m.

Question 12

A projectile is launched at a speed 10.0 m/s at angle of 20.0S1U1P1$\circ$S1S1P0 above horizontal.What is the magnitude of its acceleration at maximum height? A)0.0 m/s² B)9.40 m/s² C)9.81 m/s² D)19.8 m/s²

Accepted Answer

At the maximum height, the projectile reaches a point where its velocity becomes zero. Therefore, its acceleration at that point would be equal to the acceleration due to gravity, which is approximately 9.81 m/s^2.

Question 13

You toss a stone vertically upward and it reaches a maximum height of 2.06 m above the launch point.The initial velocity of the stone was
A)336 m.
B)215 m.
C)125 m.
D)6.36 m.

Accepted Answer

The maximum height reached by the stone can be found using the formula:

h = (v^2)/(2g)

where h is the maximum height, v is the initial velocity, and g is the acceleration due to gravity (9.8 m/s^2).

Substituting the given values, we get:

2.06 = (v^2)/(2*9.8)

Solving for v, we get:

v = sqrt(2 * 9.8 * 2.06)

v ≈ 6.36 m/s

Therefore, the initial velocity of the stone was 6.36 m/s, which corresponds to choice D.

Question 14

An arrow is shot at ground level with an initial speed of 50.0 m/s at an angle of 25.0S1U1P1$\circ$S1S1P0 above horizontal.What is its x-coordinate when the arrow is on the way up and its y coordinate is 5.00 m above the ground?
A)5.44 m
B)11.4 m
C)84.3 m
D)184 m
E)203 m

Accepted Answer

We can first use the y-coordinate information to find the time of flight. Using the equation y = viy*t + 1/2*a*t^2 and plugging in y = 5.00 m, viy = 50.0*sin(25.0°), and a = -9.80 m/s^2, we get t ≈ 3.73 s.

Next, we can use the x-coordinate information to find the horizontal distance traveled in this time. Using the equation x = vix*t and plugging in vix = 50.0*cos(25.0°) and t ≈ 3.73 s, we get x ≈ 11.4 m.

Therefore, the answer is choice B, 11.4 m.

Question 15

An airplane's speed with respect to the air is 100 m/sec.The velocity of the air with respect to the ground is 30.0 m/sec to the northeast.What is the direction of the airplane's velocity with respect to the air if the direction of its velocity with respect to the ground is due east?
A)3.00 degrees south of east
B)4.50 degrees south of east
C)12.2 degrees south of east
D)17.5 degrees south of east

Accepted Answer

To solve this problem, we need to use vector addition. Let the airplane's velocity with respect to the ground be represented by a vector pointing due east. Let the velocity of the air with respect to the ground be represented by a vector pointing to the northeast. We need to find the direction of the airplane's velocity with respect to the air, which can be represented by a vector that we'll call v.

Since we know the magnitude of both vectors (100 m/sec and 30.0 m/sec), we can use the Pythagorean theorem to find the magnitude of v:

|v|^2 = (100 m/sec)^2 + (30.0 m/sec)^2

|v| = 104.4 m/sec

Now we need to find the direction of v. We can use trigonometry for this. Let θ be the angle between v and the eastward direction. Then we have:

tan(θ) = (30.0 m/sec) / (100 m/sec)

θ = 16.7 degrees

Thus, the direction of v is 16.7 degrees south of east. However, the question asks for the answer in terms of degrees south of east, so we need to subtract θ from 180 degrees:

180 degrees - 16.7 degrees = 163.3 degrees

So the answer is 12.2 degrees south of east (since 180 - 163.3 = 16.7 and we need to round to the nearest tenth). Therefore, the correct answer is C.

Question 16

A skier leaves a ski jump with a horizontal velocity (and no vertical velocity component).If the magnitude of the vertical component of her velocity the instant before she lands is 25 m/s,how much time elapses between leaving the jump and landing?
A)2.0 s
B)1.0 s
C)4.6 s
D)2.5 s
E)Not enough information is given to determine the time of flight.

Accepted Answer

The time of flight can be determined using the vertical component of the velocity and the acceleration due to gravity. The formula $v = gt$ can be rearranged to $t = \frac{v}{g}$, where $v = 25 \, \text{m/s}$ is the final vertical velocity and $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity. Plugging in the values, $t = \frac{25}{9.8} \approx 2.55 \, \text{s}$, which rounds to 2.5 seconds.

Question 17

A skier leaves a ski jump with a horizontal velocity of 24.0 m/s (and no vertical velocity component).The skier lands two seconds later.What is the magnitude of the vertical component of the skier's displacement during the jump
A)32.2 m
B)19.6 m
C)48.0 m
D)28.3 m

Accepted Answer

Using the kinematic equation: 
Δy = vi*t + 0.5*a*t^2
where vi is the initial velocity (in the y-direction), t is the time, and a is the acceleration due to gravity (-9.8 m/s^2)
Since there is no initial vertical velocity, vi = 0
Δy = 0 + 0.5*(-9.8)*(2)^2 = -19.6 m (negative value indicates the displacement was in the downward direction)
Taking the magnitude of this value, we get 19.6 m, which is the vertical component of the skier's displacement during the jump. Therefore, the correct answer is B.

Question 18

A skier leaves a ski jump with a horizontal velocity of 24.0 m/s (and no vertical velocity component).The magnitude of the vertical component of her velocity the instant before she lands two seconds later is
A)24.0 m/s.
B)0 m/s.
C)43.6 m/s.
D)19.6 m/s.

Accepted Answer

Since the skier leaves the ski jump with no vertical velocity component, the only force acting on her during her flight is gravity, which causes her to accelerate downwards at a rate of 9.8 m/s^2. Therefore, after two seconds, her vertical velocity will have increased by 19.6 m/s (9.8 m/s^2 x 2 s). This means that the magnitude of her vertical velocity the instant before she lands will be 19.6 m/s.

Question 19

A bullet is shot horizontally from the edge of a cliff at a height of 100 m above the ground.The initial speed of the bullet is 150 m/s.It hits the ground from the base of the cliff at a horizontal distance of
A)158 m.
B)278 m.
C)498 m.
D)677 m.
E)868 m

Accepted Answer

The answer of A bullet is shot horizontally from the...

Question 20

At what angle measured with the horizon must a projectile be launched such that its range is equal to its maximum height? Ignore air resistance.
A)45.0 degrees
B)60.0 degrees
C)63.2 degrees
D)76.0 degrees

Accepted Answer

The answer of At what angle measured with the horizon...

Quiz 3: Motion in Two and Three Dimensions

An airplane moves at 120 m/sec to the west with respect to the air.The air moves at 50 m/sec to the south with respect to the ground.How fast does the airplane move with respect to the ground?

A projectile is launched at 10.0 m/s in a direction 45.0 degrees above the horizon.It hits the ground 10.0 m below its initial position.What is its speed just before hitting the ground? Ignore air resistance.

A skier leaves a ski jump with a horizontal velocity of 27.0 m/s (and no vertical velocity component) .The skier lands 2.50 seconds later.What is the magnitude of the skier's displacement during the jump

An object moves from a point with coordinates x = 0.0 m,y = -1.0 m,to a point with coordinates x = 8.0 m,y = -9.0 m in a time interval of 3.4 seconds.The magnitude of its the average velocity is

A child kicks a football with an initial velocity of 25.0 m/s at an angle of 38^$\circ$above horizontal.The ball lands at the same height from which it was kicked.What is the range of the ball?

You are walking to catch your plane and decide to take the moving sidewalk in the airport.It has an average speed of 2.0 m/s.If you are walking with a speed of 1.5 m/s and the walkway is 85.0 m long.How long will it take you to travel the 85.0 m?

A skier leaves a ski jump with a horizontal velocity (and no vertical velocity component) .If the magnitude of the horizontal component of her velocity the instant before she lands is 19.6 m/s,how much time elapses between leaving the jump and landing?

A skier leaves a ski jump with a horizontal velocity (and no vertical velocity component) .If the magnitude of the horizontal component of her velocity the instant before she lands two seconds later is 15.0 m/s,what was the horizontal component of her initial velocity

A skier leaves a ski jump with a horizontal velocity of 21.0 m/s (and no vertical velocity component) .The skier lands three seconds later.What is the magnitude of the horizontal component of the skier's displacement during the jump

If the launch speed of a kicked football was doubled,then by what factor would its hang time increase?

The maximum height above ground you can achieve with a projectile launched vertically upward from ground level,if you are able to give it an initial speed of 65.0 m/s,is

A projectile is launched at a speed 10.0 m/s at angle of 20.0^$\circ$ above horizontal.What is the magnitude of its acceleration at maximum height?

You toss a stone vertically upward and it reaches a maximum height of 2.06 m above the launch point.The initial velocity of the stone was

An arrow is shot at ground level with an initial speed of 50.0 m/s at an angle of 25.0^$\circ$ above horizontal.What is its x-coordinate when the arrow is on the way up and its y coordinate is 5.00 m above the ground?

An airplane's speed with respect to the air is 100 m/sec.The velocity of the air with respect to the ground is 30.0 m/sec to the northeast.What is the direction of the airplane's velocity with respect to the air if the direction of its velocity with respect to the ground is due east?

A skier leaves a ski jump with a horizontal velocity (and no vertical velocity component) .If the magnitude of the vertical component of her velocity the instant before she lands is 25 m/s,how much time elapses between leaving the jump and landing?

A skier leaves a ski jump with a horizontal velocity of 24.0 m/s (and no vertical velocity component) .The skier lands two seconds later.What is the magnitude of the vertical component of the skier's displacement during the jump

A skier leaves a ski jump with a horizontal velocity of 24.0 m/s (and no vertical velocity component) .The magnitude of the vertical component of her velocity the instant before she lands two seconds later is

A bullet is shot horizontally from the edge of a cliff at a height of 100 m above the ground.The initial speed of the bullet is 150 m/s.It hits the ground from the base of the cliff at a horizontal distance of

At what angle measured with the horizon must a projectile be launched such that its range is equal to its maximum height? Ignore air resistance.