Answer
The ball must be shot at a speed of $~~1.34~m/s~~$ at point $P$
Work Step by Step
We can find the time it takes the ball to fall $h_2 = 1.60~cm$:
$h_2 = \frac{1}{2}at^2$
$t = \sqrt{\frac{2h_2}{a}}$
$t = \sqrt{\frac{(2)(0.016~m)}{9.8~m/s^2}}$
$t = 0.05714~s$
We can find the required horizontal speed $v_2$ in order for the ball to travel a horizontal distance of $6.00~cm$ in this time:
$x = v_2~t$
$v_2 = \frac{x}{t}$
$v_2 = \frac{0.060~m}{0.05714~s}$
$v_2 = 1.05~m/s$
The rotational inertia of a solid ball is $I = \frac{2}{5}MR^2$
We can find a general expression for the total kinetic energy:
$K = \frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$
$K = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2$
$K = \frac{1}{2}Mv^2+\frac{1}{5}Mv^2$
$K = \frac{7}{10}Mv^2$
We can find an expression for the ball's total kinetic energy $K_2$ on the plateau:
$K_2 = \frac{7}{10}Mv_2^2$
$K_2 = \frac{7}{10}M(1.05~m/s)^2$
$K_2 = 0.772~M$
We can use conservation of energy to find the required kinetic energy $K_1$:
$K_1+U_1 = K_2+U_2$
$K_1+0 = K_2+U_2$
$K_1+0 = K_2+Mgh_1$
$K_1 = 0.772~M+(M)(9.8~m/s^2)(0.050~m)$
$K_1 = 1.26~M$
We can find the required speed $v_1$ of the ball at point $P$:
$K_1 = \frac{7}{10}Mv_1^2 = (1.26~M)$
$v_1^2 = \frac{(10)(1.26)}{7}$
$v_1 = \sqrt{\frac{12.6}{7}}$
$v_1 = 1.34~m/s$
The ball must be shot at a speed of $~~1.34~m/s~~$ at point $P$
