Vertical physics problem

Question

I feel like this should be easy and I'm making this harder than it should be.

 

A baseball player throws a baseball into the air at a height of 3.00 m above the release point, the ball is moving at 1.25 m/s. What was the ball's initial speed?

 

I don't see how I can do this without time...

Answer 1

I assume you mean that at y = 3.0 m above the release point, the speed v is 1.25 m/s. 

 

This problem involves motion at constant acceleration a = -g, so you can use the kinematic equation relating initial speed v₀ to distance traveled y and speed v at a distance y from the start point:

 

v² - v₀² = 2ay = 2(-g)y ⇒

 

v₀ = √(v² + 2gy)

 

v = 1.25 m/s (given)

v₀ = ? (what we want to solve for)

y = 3.00 m (given)

 

v₀ = √[(1.25)² + 2(9.8)(3.0)] m/s ≅ 7.77 m/s