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
Answers
Answered by
28
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...
Answered by
0
option d 0.93 is the correct answer
I PURPLE BTS
Similar questions