Answer
(a) The minimum speed is 49.5 m/s.
(b) Since the lake is 100 meters long, the rock hits the plain a distance of 50 meters from the foot of the dam.
Work Step by Step
(a) We can find the time $t$ to fall 20 meters.
$y = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2y}{g}}=\sqrt{\frac{(2)(20~m)}{9.80~m/s^2}}$
$t = 2.02~s$
We can find the minimum speed $v_0$.
$v_0 = \frac{x}{t} = \frac{100~m}{2.02~s} = 49.5~m/s$
The minimum speed is 49.5 m/s.
(b) We can find the time $t$ to fall 45 meters.
$y = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2y}{g}}=\sqrt{\frac{(2)(45~m)}{9.80~m/s^2}}$
$t = 3.03~s$
We can find the total horizontal distance $x$.
$x = v_0~t = (49.5~m/s)(3.03~s) = 150~m$
Since the lake is 100 meters long, the rock hits the plain a distance of 50 meters from the foot of the dam.
