# A stone is thrown with a horizontal velocity of 30.0 m/s from the top of a cliff 80.0 m high. How long will it take to strike the level ground at the base of the cliff, and how far from the foot of the cliff will it strike?

Dec 3, 2015

(a)

$4.04 \text{s}$

(b)

$121.2 \text{m}$

#### Explanation:

(a)

Taking the vertical component of motion we can assume that the initial velocity is zero so:

$s = \frac{1}{2} \text{g} {t}^{2}$

$\therefore t = \sqrt{\frac{2 s}{g}}$

$t = \sqrt{\frac{2 \times 80}{9.8}}$

$t = 4.04 \text{s}$

(b)

The horizontal component of velocity is constant so:

Distance = speed x time

$\therefore d = 30 \times 4.04 = 121.2 \text{m}$