Question

5) After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah's hand a distance 1.5 meters above the ground, and is moving with a speed of 22.0 m/s when it reaches a maximum height of 9.0 m above the ground.
What is the speed of the ball when it leaves Sarah's hand?

Answer 1

h(t) = -4.9t^2 + vt + 1.5
max height reached when t = v/9.8
9.0 = -4.9(v^2/4.9^2) + v(v/4.9) + 1.5
v = 7√3 = 12.1244 m/s

That is the vertical component. So, the speed is

√(7√3)^2 + 22^2 = 25.12 m/s

Answer 2

While Steve is mostly right there is an error in his algebra. The third line should read:
9.0=-4.9(v^2/9.8^2)+v(v/9.8)+1.5.
The rest can be correctly solved using this equation.

Answer 3

2*a(x-x0)=(v^2-v0^2)
2*-9.8(9-1.5)=(v^2-(22*22))
v=18.35

Answer 4

Where does the t=v/9.8 come form