Answer to Question #80735 in Physics for J

Question

A boy throws a rock with an initial velocity of 22 m/s at 28 ° above the horizontal. How long does it take for the rock to reach the maximum height of its trajectory if air resistance is negligibly small and g = 9.80 m/s2?

Answer 1

Let’s first find the vertical component of the rock’s velocity:
v_0y=v_0 sinθ=22 m/s∙sin〖28〗^°=10.33 m/s.
We can find the time that the rock needs to reach the maximum height of its trajectory from the kinematic equation:
v_y=v_0y-gt,
here, v_y=0 is the velocity of the rock at the maximum height, v_0y is the vertical component of the rock’s velocity, g=9.8 m⁄s^2 is the acceleration due to gravity and t is the time that the rock needs to reach the maximum height of its trajectory.
Then, we get:
0=v_0y-gt,
t=v_0y/g=(10.33 m/s)/(9.8 m/s^2 )=1.05 s.