A projectile is launched from the edge of a cliff with a velocity of 15.0 m/s at an angle of 30.0 degrees above the horizontal. Ignoring air resistance, what is the velocity of the projectile 2.00 seconds later?
A) 17.8 m/s at an angle of 38 degrees above the horizontal
B) 11.4 m/s at an angle of 43 degrees below the horizontal
C) 17.8 m/s at an angle of 43 degrees below the horizontal
D) 11.4 m/s at an angle of 38 degrees below the horizontal
E) 17.8 m/s at an angle of 43 degrees above the horizontal

Question

Quizplus · Accepted Answer

First, we will separate the initial velocity into its x and y components.

Vx = 15.0 m/s cos(30.0°) = 13.0 m/s 
Vy = 15.0 m/s sin(30.0°) = 7.50 m/s

Using these initial velocities and ignoring air resistance, we can find the final velocity at 2.00 seconds using the following kinematic equations:

x = x0 + (Vx)t 
y = y0 + (Vy)t - (1/2)gt^2 
Vx = Vx0 
Vy = Vy0 - gt

We know that at 2.00 seconds: 
t = 2.00 s 
Vx0 = 13.0 m/s 
Vy0 = 7.50 m/s

We can find the final x position of the projectile using the first kinematic equation:

x = x0 + (Vx)t 
x = 0 + (13.0 m/s)(2.00 s) = 26.0 m

Next, we can find the final y position of the projectile using the second kinematic equation:

y = y0 + (Vy)t - (1/2)gt^2 
y = 0 + (7.50 m/s)(2.00 s) - (1/2)(9.81 m/s^2)(2.00 s)^2 
y = 7.50 m

Finally, we can find the final velocity of the projectile using the third kinematic equation:

Vx = Vx0 
Vy = Vy0 - gt 
Vx = 13.0 m/s 
Vy = 7.50 m/s - (9.81 m/s^2)(2.00 s) = -12.6 m/s

Combining the final x and y velocities using the Pythagorean theorem, we get:

V = sqrt(Vx^2 + Vy^2) 
V = sqrt((13.0 m/s)^2 + (-12.6 m/s)^2) = 17.8 m/s

To find the angle of the final velocity, we can use:

theta = tan^-1(Vy/Vx) 
theta = tan^-1(-12.6 m/s/13.0 m/s) = -43°

Since the angle is negative, we know that the velocity is below the horizontal axis. Therefore, the answer is (C) 17.8 m/s at an angle of 43 degrees below the horizontal.

A Projectile Is Launched from the Edge of a Cliff