Question

An ball rolls off a cliff 21.0 m above the ground with an initial horizontal speed of 5.0 m/s. Acceleration due to gravity is 9.8 m/s2.
a) How long after leaving the top of the cliff does the ball hit the ground?

b) How far from the base of the cliff does the ball land?

Answer 1

Question #54122, Physics / Mechanics | Kinematics | Dynamics |

A ball rolls off a cliff $21.0\,\mathrm{m}$ above the ground with an initial horizontal speed of $5.0\,\mathrm{m/s}$. Acceleration due to gravity is $9.8\,\mathrm{m/s^2}$.

a) How long after leaving the top of the cliff does the ball hit the ground?

b) How far from the base of the cliff does the ball land?

Solution:

a) The vertical motion of the object is described by the equation:

$h = v_0 t + 1/2 \, \text{gt}^2, v_0$ – the initial vertical speed, which equals zero, $h$ – the height of the cliff.

Therefore,

$$h = 1/2 \, \text{gt}^2,$$ $$t = (2h/g)^{1/2} = (42 \, \text{m}/9.8 \, \text{m} \, \text{s}^{-2})^{1/2} = 2.07 \, \text{s}$$

Answer (a): The object hits the ground in 2.07 s.

b) The horizontal motion with the constant speed is defined by the equation:

$d = v_0 t$, where $d$ – the distance between the base of the cliff and a landing place, $t$ – the time of the object flight.

$$d = 5.0 \, \text{m/s} \times 2.07 \, \text{s} = 10.35 \, \text{m}$$

Answer (b): The ball moves 10.35 m horizontally.

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