Question

A batter hits a fly ball which leaves the bat 0.89 m above the ground at an angle of 62 ∘ with an initial speed of 28 m/s heading toward centerfield. Part A) Ignore air resistance.How far from home plate would the ball land if not caught? Part B) The ball is caught by the centerfielder who, starting at a distance of 103 m from home plate, runs straight toward home plate at a constant speed and makes the catch at ground level. Find his speed.

Answer 1

First of all determine the horiontal and vertical components of theinitial speed of the ball.

Vx(0)=28*cos(62)
Vy(0)=28*sin(62)

(a) Now use the following expression -  

y(t)=0.89 + Vy(0)*t - 0.5*9.81*t^2

For the ball to be caught, we have -  
y(t)=0

means-
0.89 + 28*sin(62)*t-0.5*9.81*t^2 = 0

=> 0.89 + 24.7t - 4.9t^2 = 0

=> 4.9t^2 - 24.7t - 0.89 = 0

=> t = [24.7 + sqrt(24.7^2 + 4*4.9*0.89)] / (2*4.9) = [24.7 + 25.1] / 9.8 = 5.08 s

discard the negative sign.
So, without being caught, the ball will traverse a distance of
x(t) = Vx(0)*t
= 28*cos(62)*5.08
= 66.78 m

(b) The center fielder must run 103 - 66.78 = 36.22 m in 5.08 s to catch the ball at the ground level.

So, the requisite speed = 36.22 / 5.08 = 7.13 m/s.