Question

A tennis ball is thrown from ground level with velocity directed 30° above the horizontal. if it takes the ball 0.5 s to reach the top of its trajectory, what is the magnitude of the initial velocity?

Answer 1

The motion of the tennis ball on the vertical axis is an uniformly accelerated motion, with deceleration of $g=-9.81 m/s^2$ (gravitational acceleration).

The component of the velocity on the y-axis is given by the following law:
$v_y(t) = v_{y0}+gt$
At the time t=0.5 s, the ball reaches its maximum height, and when this happens, the vertical velocity is zero (because it is a parabolic motion): $v_y(0.5 s)=0$. Substituing into the previous equation, we find the initial value of the vertical component of the velocity:
$v_{y0}=-gt=-(-9.81 m/s^2)(0.5 s)=4.9 m/s$

However, this is not the final answer. In fact, the ball starts its trajectory with an angle of $30^{\circ}$. This means that the vertical component of the initial velocity is
$v_{y0}=v_0 sin 30^{\circ}$
We found before $v_{0y}=4.9 m/s$, so we can substitute to find $v_0$, the initial speed of the ball:
$v_0 = \frac{v_{y0}}{sin 30^{\circ}}=9.81 m/s$