Question

A baseball pitcher releases the ball at a height of 1.9 m above the ground at a horizontal distance of 18.5 m from home plate. The initial angle is 0.50° above the horizontal. In order to throw a strike, the ball must cross home plate between 55 and 120 cm above the ground. For what range of initial speeds will the pitch be in the strike zone? Ignore the effects of air resistance.

Answer 1

From the given figure,

initial height=1.9 m

final height=0.55m

change in height= (1.9-0.55)m= 1.35 m

horizontal distance= 18.5 m

let initial speed= u

initial angle= 0.5 degrees

18.5= ucos0.5 * t

t=18.5/(ucos0.5)................................(1)

-1.35=usin0.5t - (1/2)(9.8)t^2............................(2)

replacing (1) in (2)

-1.35=usin0.5*[18.5/(ucos0.5)] - (1/2)(9.8)*[18.5/(ucos0.5]^2

solving we get

u₁=10.27 m/s

Similarly for second height,

change in height= (1.9-1.2)m= 0.7 m

-0.7=usin0.5*[18.5/(ucos0.5)] - (1/2)(9.8)*[18.5/(ucos0.5]^2

u₂=10.89 m/s

The range of initial speeds are 10.27 to 10.89 m/s