(a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it?

Question

situation: The fastest measured pitched baseball left the pitcher's hand at a speed of 45.0 m/s. If the pitcher was in contact with the ball over a distance of 1.50 m and produced constant acceleration,

Answer 1

1)just use this :V^2-V'^2=2ad

V is 45m/s ans V' is 0 because at the first it doesn't have any speed and also d is 1.50 m so u can find a:

45^2=2x1.5xa and then a=675 m/s^2

2)for finding t use this: V=at+V'

45=675t+0 then t=45/675=0.06666666

Answer 2

2ad = Vf^2 - Vi^2

a = (Vf^2 - Vi^2)/2d

a = (45.0^2 - 0^2)/(2)(1.50)

a = 675 m/s^2   ANSWER

a = (Vf - Vi)/t

t = (45.0 - 0)/675

t = .067 sec    ANSWER

Hope this helps.

teddy boy

Answer 3

x = 1/2 a t^2 where x = 1.5 m

so that

1.5 = 1/2 a t^2

and

v = at where v = 45 m/s

so that

45 = a t

Now this is the same time, t, so we can solve these for a.

from the second we have t = 45/a

substituting this into the first we have:

1.5 = 1/2 a (45/a)^2

1.5 = .5( 2,025/a)

a = 1,012.5/1.5

a = 675 m/s^2

This is the acceleration he gave the ball.

Now put this value for a into 45 = at to get:

45 = 675t or t = 2/30 or 0.0666666666666667 seconds.

that's it!