Question

A baseball pitcher throws a ball with a speed of 41 m/s. Estimate the average acceleration of the ball during the throwing motion. In throwing the baseball, the pitcher accelerates the ball through a displacement of about 3.5 m, from behind the body to the point where it is released.

Answer 1

v= u+at
u= 0 ,v= 41m/s
s= 1/2 a t^2
t= sqrt(2s/a)
given s (distance)= 3.5m
41= a*t
41= a* sqrt(2s/a)
41= sqrt(2as)
1681 = 2*3.5 *a
a= 24.14m/s^2

Answer 2

a= 240.14m/s^2

Answer 3

The concepts that are can be used in this problem are:
v = v0 + at
x = x0 + v0t + 1/2(at^2)
v^2 = V0^2 + 2a(x - x0)

v - velocity
a-constant acceleration

t - time
x - distance

Answer 4

How did you determine which concept to use?

Answer 5

Use the 3rd equation. V(final) is 41m/s, v(initial) is 0m/s because the baseball was accelerated from rest. The quantity (x-x0) is the displacement, 3.5m.

Solving for a yields:
a=(Vf^2 - Vi^2)/(2(x-x0)=(41^2-0)/(2*3.5)=240m/s^2
:)