Question

If the ball shown in the figure lands in 0.5 s, about what height was it thrown from?

Answer 1

Answer: 1.22 m

Explanation:

The equation of motion in this situation is:

$y=y_{o}+V_{oy}t-\frac{g}{2}t^{2}$ (1)

Where:

$y=0$ is the final height of the ball

$y_{o}=h$ is the initial height of the ball

$V_{oy}=V_{o}sin(0\°)=0$ is the vertical component of the initial velocity (assuming the ball was thrown vertically and there is no horizontal velocity)

$t=0.5 s$ is the time at which the ball lands

$g=9.8 m/s^{2}$ is the acceleration due gravity

So, with these conditions the equation is rewritten as:

$h=\frac{g}{2}t^{2}$ (2)

$h=\frac{9.8 m/s^{2}}{2}(0.5 s)^{2}$ (3)

Finally:

$h=1.22 m$