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:
$E_1 = \frac{1}{2} mv_1^2 + mgh_1$

Total energy at the roof:
 $E_2 = \frac{1}{2} mv_2^2 + mgh_2$

Energy must be conserved so :
$E_1 = E_2 \\ \\ \frac{1}{2} mv_1^2 + mgh_1 = \frac{1}{2} mv_2^2 + mgh_2 \\ \\ \frac{1}{2} m (v_1^2 - v_2^2) = mg(h_2 - h_1) \\ \\ \frac{1}{2} (v_1^2 - v_2^2) = g(h_2 - h_1) \\ \\ v_1^2 - v_2^2 = 2g(h_2 - h_1)$

If v₁ is the initial velocity and v₂ is the final velocity, which is zero at the roof:
$v_1^2 = 2g(h_2 - h_1) \\ \\ v_1 = \sqrt{2g(h_2 - h_1)}$

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