91Ó°ÊÓ

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, $\mathrm{Δ}y=\left(V_{iy}\right)\left(t\right)+\frac{1}{2}a{t}^2$ …(¾±)

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:

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

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$}\right.\right)×\left(0.5  \text{s}\right)\\ & =& 8.85\text{m}\end{array}$

Hence, the distance from the fence is 8.85 m.