91Ó°ÊÓ

The range is the horizontal distance covered by the projectile in its flight.

The initial velocity with which the ball is launched,$v_0=28\mathrm{m}/\mathrm{s}$

The angle at which the ball is launched from the bat,$\mathrm{θ}=61°$

The initial height of the ball at the time of launch,$y_0=0.90\mathrm{m}$

Assume that the ball lands at a horizontal distance R and the speed of the centerfield is V.

The initial velocity in the vertical direction:

$v_{\mathrm{y}}=v_0\sin \mathrm{θ}$

For motion in the vertical direction, you can write the second equation of motion as:

$0-y_0=v_0\sin \mathrm{θ}t+\frac{1}{2}g{t}^2$

(Here, t is the time of flight)

Substituting the values in the above equation, you get:

$0-0.90=28\sin 61°t+\frac{1}{2}\left(-9.8\right)t^2$

From the above equation, you get:

$4.9t^2-24.49t-0.90=0$

Solving the above quadratic, you get:

t = 5.03 s (Reject the negative root as time cannot be negative)

Calculating the horizontal range

The initial velocity along the horizontal direction:

$v_{\mathrm{x}}=v_0\cos \mathrm{θ}$

Using the second equation of motion for horizontal motion, you get:

$R=v_0\cos \mathrm{θ}t+0$

(As acceleration along the horizontal direction is zero)

Substituting the values in the above equation, you get:

$\begin{array}{c}R=28\cos 61°×5.03\\ =68.28\mathrm{m}\end{array}$

The distance that the centerfield has to travel for catching the ball:

$\begin{array}{c}d=105-68.28\\ =36.72\mathrm{m}\end{array}$

The speed of the centerfield can be calculated as:

$V=\frac{d}{t}$

(As the time taken by the centerfield is equal to the time of flight of the ball)

Substituting the values in the above equation, you get:

$\begin{array}{c}V=\frac{36.72}{5.03}\\ =7.3\mathrm{m}/\mathrm{s}\end{array}$

The velocity of the centerfield is $7.3\mathrm{m}/\mathrm{s}$.