Question

After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it reaches Julie's horizontal position. Assume the ball leaves Sarah's hand a distance 1.5 meters above the ground, reaches a maximum height of 8 m above the ground, and takes 1.619 s to get directly over Julie's head. What is the speed of the ball when it leaves Sarah's hand?

Answer 1

Answer:

Vo=20.2m/s

Explanation:

We can calculate both components of initial velocity and the use them to calculate the speed as $\sqrt{V_{ox}^{2}+V_{oy}^{2}}$

From the vertical movement to the maximum height:

$V_{fy}=V_{oy}-g*t$  since $V_{fy}=0$

$V_{oy}=g*t=16.19m/s$

Now, from the horizontal movement, also to the point of maximum height:

$Y_{f}=Y{o}+V_{oy}*t-\frac{g*t^{2}}{2}$  Solving for $V_{oy}$

$V_{oy}=12.1m/s$

Finally we calculate the magnitude of the new vector:

$V_{o}=\sqrt{V_{ox}^{2}+V_{oy}^{2}}=20.2m/s$