A projectile is projected from the origin with a velocity of 20.0 m/s at an angle of 30.0 degrees above the horizontal.The projectile lands on a roof that is 4.00 m height.What is the speed when the projectile hits the roof?
A) 19.2 m/s
B) 17.9 m/s
C) 12.5 m/s
D) 9.20 m/s
E) 8.40 m/s

Question

Quizplus · Accepted Answer

First, we need to find the time it takes for the projectile to reach the roof. We can use the equation:
$$y=y_0+v_{0y}t+\frac{1}{2}at^2$$
where $y$ is the final height of the projectile, $y_0$ is the initial height (zero in this case), $v_{0y}$ is the initial vertical velocity (calculated from the initial velocity and launch angle), $a$ is the acceleration due to gravity (-9.81 m/s$^2$), and $t$ is the time. Solving for $t$:
$$4.00	ext{ m} = 0 + (20.0 	ext{ m/s} \sin 30.0^\circ)t + \frac{1}{2}(-9.81	ext{ m/s}^2)t^2$$
$$t^2 - 4.08t + 0.816 = 0$$
Using the quadratic formula, we get $t=0.200$ s or $t=3.88$ s. We can discard the longer time as it would indicate that the projectile goes up higher than the roof (which is not possible).
Next, we can use the horizontal component of the initial velocity to find the distance to the roof:
$$x=v_{0x}t=20.0	ext{ m/s}\cos 30.0^\circ \cdot 0.200 	ext{ s} = 3.46	ext{ m}$$
Finally, we can use the equation for the final velocity $v_f$ in terms of initial velocity $v_0$, acceleration $a$, and distance $d$:
$$v_f^2 = v_0^2 + 2ad$$
We can solve for $v_f$:
$$v_f = \sqrt{v_0^2 + 2ad} = \sqrt{(20.0	ext{ m/s})^2 + 2(-9.81	ext{ m/s}^2)(4.00	ext{ m}-0)} = 17.9	ext{ m/s}$$
Therefore, the answer is (B).

A Projectile Is Projected from the Origin with a Velocity