A baseball player hits a ball straight up into the air. It leaves the bat with a speed of 120 km/h. In the absence of air resistance, how fast would the ball be traveling when it is caught at the same height above the ground as it left the bat? Explain.

When the acceleration is constant, in that case, we use the third equation of motion to determine the initial and final velocity or the displacement of the object.

Given data:

The initial velocity of the ball is $u=120\frac{\mathrm{km}}{\mathrm{h}}$.

Initially, the ball is going upward, so the acceleration of the ball will be $g=-9.81\frac{\mathrm{m}}{{\mathrm{s}}^2}$.

At the maximum height, the final velocity of the ball will be $v=0\frac{\mathrm{m}}{\mathrm{s}}$.

The height of the ball above the ground can be calculated as:

$\begin{array}{c}s=\frac{v^2-u^2}{2g}\\ s=\frac{{\left(0\right)}^2-{\left(120\frac{\mathrm{km}}{\mathrm{h}}\times \frac{1\frac{\mathrm{m}}{\mathrm{s}}}{3.6\frac{\mathrm{km}}{\mathrm{h}}}\right)}^2}{\left(2\right)\left(-9.81\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)}\\ s=56.7\mathrm{m}\end{array}$

When the ball is $56.7\mathrm{m}$above the hit point, the initial speed of the ball at that time will be $u=0\frac{\mathrm{m}}{\mathrm{s}}$.

The speed of the ball when it is caught at the same height can be calculated as:

$\begin{array}{c}s=\frac{v^2-u^2}{2g}\\ \left(56.7\mathrm{m}\right)=\frac{v^2-{\left(0\right)}^2}{2\left(9.81\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)}\\ v=\left(33.35\frac{\mathrm{m}}{\mathrm{s}}\times \frac{3.6\frac{\mathrm{km}}{\mathrm{h}}}{1\frac{\mathrm{m}}{\mathrm{s}}}\right)\\ v=120\frac{\mathrm{km}}{\mathrm{h}}\end{array}$

Thus, the speed of the ball when it is caught at the same height is $120\frac{\mathrm{km}}{\mathrm{h}}$.

Hence, the speed of the ball is the same but opposite in direction.