Answer
At the base of the hill, the block must have a minimum speed of 42.0 m/s.
Work Step by Step
We can find the time it takes the block to fall from a height of 70 meters down to a height of 50 meters.
$y = \frac{1}{2}gt^2$
$t^2 = \frac{2y}{g}$
$t = \sqrt{\frac{2y}{g}}$
$t = \sqrt{\frac{(2)(20~m)}{9.80~m/s^2}}$
$t = 2.02~s$
We can find the horizontal speed required to travel 40 meters in this time.
$v = \frac{d}{t} = \frac{40~m}{2.02~s}$
$v = 19.8~m/s$
We can find the required speed at the base of the hill.
$K_1+U_1 = K_2+U_2$
$\frac{1}{2}mv_1^2+0 = \frac{1}{2}mv_2^2+mgh$
$v_1^2 = v_2^2+2gh$
$v_1 = \sqrt{v_2^2+2gh}$
$v_1 = \sqrt{(19.8~m/s)^2+(2)(9.80~m/s^2)(70~m)}$
$v_1 = 42.0~m/s$
At the base of the hill, the block must have a minimum speed of 42.0 m/s.
