A batter hits a pitched ball when the center of the ball is 1.22 mabove the ground. The ball leaves the bat at an angle of $\mathbf{45}\mathbf{°}$with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of 107 m.(a) Does the ball clear a 7.32 m-high fence that is 97.5 mhorizontally from the launch point? (b) At the fence, what is the distance between the fence top and the ball center?

It is given that, at the fence, the ball is at a height of 1.22 m above the ground,

the angle of projection is$\theta =45°$

The range of projectile is$R=107\mathrm{m}$and

the horizontal distance is$x=97.5\mathrm{m}$

As ball is hit above the ground, take the point of fence as upside to ground.Horizontal range of the ball and the initial angle at which the ball is projected are given. So,kinematic equations can be used to find the initial velocity and time of the ball to reach a certain horizontal distance.

Required formulae are as follow:

The horizontal distance in kinematic equation can be written as,

$x=v_{0}t+\frac{1}{2}a{t}^2$ (i)

The range of the projectile is given by,

$R=\frac{v_0^2\sin 2\theta }{g}$ (ii)

Let’s take the origin as the point of impact. To find the height of the ball at a given horizontal distance, find its initial velocity and time to reach that given horizontal distance.

Using equation (ii),

$$\begin{gathered} R=\frac{v_0^2\sin 2\theta }{g} \\ {v}_0=\sqrt{\frac{Rg}{\sin 2\theta }} \\ {v}_0=32.38\mathrm{m}/\mathrm{s} \end{gathered}$$

Using equation (i),

$x=v_{0}t+\frac{1}{2}a{t}^2$

But, the time for horizontal distance and no acceleration in horizontal direction are needed.

So,

$$\begin{gathered} x=v_0\cos \theta \times t \\ t=\frac{x}{v_0\cos \theta } \\ t=4.258\mathrm{s} \end{gathered}$$

Now initial velocity and time to reach horizontal distance as well as the angle of projection are obtained. So by using 2^nd kinematic equation, find the height of ball for that situation.

$$\begin{gathered} y=\left(v_0\sin \theta \right)t+\frac{1}{2}a{t}^2 \\ y=97.53-88.92 \\ y=8.61\mathrm{m} \end{gathered}$$

From the point of fence,

$$\begin{gathered} ∆y=8.61+1.22 \\ =9.83\mathrm{m} \end{gathered}$$

So,theball is at 9.383 m height from the ground at 97.5 m horizontal distance.

And at 4.26 sec , the ball is $9.83-7.32=2.51\mathrm{m}$above the fence.

Therefore, using the given conditions, the initial velocity oftheball can be found, and from this,thetime taken bytheball to reach some horizontal distance can be found. From all this information, the height of ball at that particular situation can be found.

Kinematic equations and equation of range of projectile, which is generally obtained from kinematic equations plays a main role in all the calculations. First of all, the value of initial velocity by using a range equation is obtained and that value will be helpful for finding the time of the ball to reach a certain horizontal distance.