Question

A baseball pitcher throws a ball at 40 ms^-1. if the acceleration is approximately constant over a distance of 2 m, how large is it?

Answer 1

We have the equation of motion
`v^2=u^2+2as`, where v i the final velocity, u is the initial velocity, a is the acceleration and s is the displacement

Here final velocity, v = 40m/s

Initial velocity, u = 0 m/s

Displacement s = 2 m

Substituting
`40^2=0^2+2*a*2\\ \\ a=400m/s^2`

So the baseball pitcher accelerates at 400m/
`s^2` to release a ball at 40 m/s.