Physics problem,Would appreciate hints, advices or guide how to solve this problem? Thank you?

Question

A major-league pitcher can throw a ball in excess of 40.5 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is 17.0 m away from the point of release?

Answer 1

Consider the horizontal and the vertical motions separately.

Horizontal motion :

v = 40.5 m/s

s = 17

t = s/v = 0.42

Vertical motion :

u=0 m/s

t = 0.42 s

a = 9.8 ms/^2

s = ut + 1/2*at^2 = 0.86 m

Hope this helps.

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Answer 2

divide the ball trajectory into horizontal and vertical path ways.

Horizontally:

speed=40.5              distance=17

time=t=?

so time= 17/40.5=0.42 s

Vertically:

initial velocity=u=0                  acceleration(a)=10

displacement(S)=?                  time(t)=0.42

S=ut+0.5at^2

  =0.5*10*0.42^2

  =0.88 m below the releasing point level.