A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 safter being hit. Then 2.5 s after reaching its maximum height, the ball barely clears a fence that is 97.5 mfrom where it was hit. Assume the ground is level. (a) What maximum height above ground level is reached by the ball? (b) How high is the fence? (c) How far beyond the fence does the ball strike the ground?

1. The time taken by the ball to reach the maximum height is 3.0 s
2. The time taken by the ball to clear the fence which is at from y = 97.5 m the hitting point is 2.5 s.

Using the kinematic equations, we can find the maximum height reached by an object and the horizontal distance covered in that time interval.

Formulae:

The second equation of kinematic motion, $\Delta y=\left(V_{iy}\right)\left(t\right)+\frac{1}{2}a{t}^2$ …(i)

Diagram:

After hitting the ground, the ball reaches to the maximum height (x₁, y₁) in 3 seconds.

Using the Newton’s second kinematic equation, we can get the maximum height reached by the ball as follows:

$\begin{array}{rcl}0-y_1& =& 0-\frac{1}{2}g{t_1}^2\\ y_1& =& \frac{1}{2}\left(9.8 {\text{m/s}}^2\right){\left(3 \text{s}\right)}^2\\ & =& 44.1\text{m}\\ & & \end{array}$

So, the maximum height above ground reached by the ball is 44.1 m.

The height of the fence can be found using equation (i) as:

$\begin{array}{rcl}{y}_2-y_1& =& 0-\frac{1}{2}g{\left(t_2\right)}^2\\ y_2& =& \left(44.1 \text{m}\right)-\frac{1}{2}\left(9.8 {\text{m/s}}^2\right){\left(2.5 \text{s}\right)}^2\\ & =& 13.48\text{m}\\ & & \end{array}$

Hence, the value of the height is 13.48.

The horizontal displacement of the ball when it strikes the fence is given as:

$\begin{array}{rcl}97.5 \text{m}& =& V_{ix}\left(5.5 \text{s}\right)\\ V_{ix}& =& 17.7\text{m/s}\\ & & \end{array}$

The total time of flight of the ball using the projectile concept is given as:

$\begin{array}{rcl}2\times t& =& 2\times 6 \text{s}\\ & =& 12 \text{s}\\ & & \end{array}$

The total time in which ball reaches at the ground from the maximum height is 6 seconds and in 5.5 seconds it strikes the fence. So, the time taken by it to reach the ground from the point where it strikes the fence is given by ,$6-5.5=0.5\text{s}$.

Thus, the distance from the fence where the ball strikes the ground is given using the above time value as:

$\begin{array}{rcl}x& =& \left(17.7\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)\times \left(0.5 \text{s}\right)\\ & =& 8.85\text{m}\end{array}$

Hence, the distance from the fence is 8.85 m.