Answer to Question #141014 in Physics for Jeanly

Question

A ball is thrown horizontally with a velocity of 20 m/s from a cliff of height 50.0 m.
(a) What is the speed of the ball after 2.0 s?
(b) What is the time taken for the ball to reach the bottom of the cliff?

Answer 1

a) The horizontal component of the velocity is:

$$v_x = 20m/s$$

at any moment of time.

The vertical component:

$$v_y = gt = 9.81m/s^2\cdot 2s = 19.62m/s$$

here $g = 9.81m/s^2$ is the gravitational acceleration and $t = 2s$ is the time ellapsed.

The total speed is:

$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 19.62^2} = 28.02\space m/s$$

b) The distance travelled by the ball is:

$$h = \dfrac{gt^2}{2}$$

where $h = 50m$ is the height of the cliff and $t$ is the time taken for the ball to reach the bottom of the cliff. Expressing $t$, obtain:

$$t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2\cdot 50}{9.81}} \approx 3.19s$$

Answer. a) 28.02 m/s, b) 3.19 s.