Question

A pitcher throws his fastball horizontally at
42.1 meters per second. How far does it drop
before crossing the plate, 18.3 meters away
a. 0.8 m
b. 1.2 m
c. 2.2 m
d. 0.93 m​

Answer 1

Welcome Prashasti,

◆ Answer -

(d) 0.93 m

● Explanation -

In horizontal direction, ux = 42.1 m/s, sx = 18.3 m,

t = sx / ux

t = 18.3 / 42.1

t = 0.4347 s

In vertical direction, uy = 0 m/s, t = 0.4347 s,

sy = uy.t + 1/2 gt^2

sy = 0 × 0.4347 + 1/2 × 9.8 × 0.4347^2

sy = 4.9 × 0.189

sy = 0.9258 m

Therefore, the ball have to drop 0.9258 m long before crossing the plate.

* Note here that given initial velocity is purely horizontal while initial vertical velocity is zero.

Thanks for asking...

Answer 2

option d 0.93 is the correct answer

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