Solve the problem. Unless stated otherwise, assume that the projectile flight is ideal, that the launch angle is measured from the horizontal, and that the projectile is launched from the origin over a horizontal surface
-A projectile is fired with an initial speed of $512 \mathrm {~m} / \mathrm { sec }$ at an angle of $45 ^ { \circ }$. What is the greatest height reached by the projectile? Round your answer to the nearest tenth.
A) 26,749.4 m
B) 6687.3 m
C) 73.9 m
D) 65,536.0 m

Question

Quizplus · Accepted Answer

The initial velocity can be resolved into horizontal and vertical components: 
\begin{align*}
v_{0,x} &= v_0 \cos 	heta = 512 \cos 45^\circ = 362.04 \mathrm{~m/s}\
v_{0,y} &= v_0 \sin 	heta = 512 \sin 45^\circ = 362.04 \mathrm{~m/s}
\end{align*}
At the highest point of the trajectory, the vertical velocity is zero. Using the kinematic equation $v^2 = v_0^2 + 2ad$, we can find the displacement in the y-direction: 
\begin{align*}
v_y^2 &= v_{0,y}^2 + 2ad_y\
0 &= (512\sin 45^\circ)^2 - 2 \cdot 9.8 \cdot d_y\
d_y &= \frac{(512\sin 45^\circ)^2}{2 \cdot 9.8} \approx 6687.3 \mathrm{~m}
\end{align*}
Therefore, the greatest height reached by the projectile is approximately 6687.3 m.

Solve the Problem 512 m/sec512 \mathrm {~m} / \mathrm { sec }512 m/sec

Solve the Problem $512 \mathrm {~m} / \mathrm { sec }$