Question

A batter hits a fly ball which leaves the bat 0.90 m above the ground at an angle of${61}^o$with an initial speed of$28\mathrm{m}/\mathrm{s}$heading toward centerfield. Ignore air resistance. (a) How far from home plate would the ball land if not caught? (b) The ball is caught by the centerfielder who, starting at a distance of 105 m from home plate just as the ball was hit, runs straight towards home plate at a constant speed and makes the catch at ground level. Find his speed.

Short answer

(a) The ball will land at a horizontal distance of 68.28 m.

(b) The speed of the centerfield will be $7.3\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Step 1. Understanding the concept of range

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

Step 2. Given data and assumptions

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,$\theta =61°$

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.

Step 3. Calculating the horizontal range and the time of flight

The initial velocity in the vertical direction:

$v_{\mathrm{y}}=v_0\sin \theta$

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

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

(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:

$v_{\mathrm{x}}=v_0\cos \theta$

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

$R=v_0\cos \theta t+0$

(As acceleration along the horizontal direction is zero)

Substituting the values in the above equation, you get:

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

Step 4. Calculating the speed of the centerfield

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 $7.3\mathrm{m}/\mathrm{s}$.