Answer
(a) The minimum initial speed 12.9 m/s
(b) The speed when the ball hits the ground is 14 m/s
Work Step by Step
(a) Let $KE_1$ be the initial kinetic energy. Let $PE_1$ be the potential energy at height $h_1=1.5~m$. Let $KE_2$ be the kinetic energy at maximum height. Let $PE_2$ be the potential energy at maximum height $h_2 = 10~m$.
We can use conservation of energy to find the minimum speed $v_1$.
$KE_1+PE_1 = KE_2+PE_2$
$\frac{1}{2}mv_1^2+mg~h_1 = 0+mg~h_2$
$v_1^2 = 2g~(h_2-h_1)$
$v_1 = \sqrt{2g~(h_2-h_1)}$
$v_1 = \sqrt{(2)(9.80~m/s^2)(10~m-1.5~m)}$
$v_1 = 12.9~m/s$
The minimum initial speed 12.9 m/s
(b) Let $KE_3$ be the kinetic energy when the ball hits the ground. Let $PE_3$ be the potential energy at $h_3 = 0$.
We can use conservation of energy to find the speed $v_3$ when the ball hits the ground.
$KE_3+PE_3 = KE_2+PE_2$
$\frac{1}{2}mv_3^2+0 = 0+mg~h_2$
$v_3^2 = 2g~h_2$
$v_3 = \sqrt{2g~h_2}$
$v_3 = \sqrt{(2)(9.80~m/s^2)(10~m)}$
$v_3 = 14~m/s$
The speed when the ball hits the ground is 14 m/s
