Question

With what minimum speed must you toss a 200 g ball straight up to just touch the 15-m-high roof of the gymnasium if you release the ball 1.6 m above the ground? solve this problem using energy.

Answer 1

Total energy before the toss:

`E_1 = (1)/(2) mv_1^2 + mgh_1`

Total energy at the roof:

`E_2 = (1)/(2) mv_2^2 + mgh_2`

Energy must be conserved so :

`E_1 = E_2 \\ \\ (1)/(2) mv_1^2 + mgh_1 = (1)/(2) mv_2^2 + mgh_2 \\ \\ (1)/(2) m (v_1^2 - v_2^2) = mg(h_2 - h_1) \\ \\ (1)/(2) (v_1^2 - v_2^2) = g(h_2 - h_1) \\ \\ v_1^2 - v_2^2 = 2g(h_2 - h_1)`

If v₁ is the initial velocity and v₂ is the final velocity, which is zero at the roof:

`v_1^2 = 2g(h_2 - h_1) \\ \\ v_1 = √(2g(h_2 - h_1))`


`v_1 = \sqrt{(2)(9.81 (m)/(s^2) )(15m - 1.6m)} = 16,2 (m)/(s)`