Question

A stone is thrown at an angle of 30 degrees above the horizontal from the top edge of a cliff with an initial speed of 12 m/s. A stop watch measures the stone’s trajectory time from top of cliff to bottom to be 4.8s. What is the height of the cliff?

Answer 1

v = initial velocity of launch of the stone = 12 m/s

θ = angle of the velocity from the horizontal = 30

Consider the motion of the stone along the vertical direction taking upward direction as positive and down direction as negative.

v₀ = initial velocity along vertical direction = v Sinθ = 12 Sin30 = 6 m/s

a = acceleration of the stone = - 9.8 m/s²

t = time of travel = 4.8 s

Y = vertical displacement of stone = vertical height of the cliff = ?

using the kinematics equation

Y = v₀ t + (0.5) a t²

inserting the values

Y = 6 (4.8) + (0.5) (- 9.8) (4.8)²

Y = - 84.1 m

hence the height of the cliff comes out to be 84.1 m

Answer 2

The height of the cliff if the stone’s trajectory time from the top of the cliff to bottom to be 4.8s is 84.06m

To get the height of the cliff, we will have to use Newton's second law of motion as shown below;

$S=ut+\frac{1}{2}at^2$

Since the stone is thrown upwards, a = -g, the formula will become;

$S=ut-\frac{1}{2}gt^2$

Get the initial vertical velocity "u"

$u = u_0sin \theta\\u=12sin30^0\\u=6.0m/s$

Given the following parameters

t = 4.8secs

g = 9.8m/s²

Substitute the given parameters into the formula as shown:

$S=6(4.8)-0.5(9.8)(4.8)^2\\S=28.8-112.896\\S = -84.06m$

This shows that the height of the cliff if the stone’s trajectory time from the top of the cliff to bottom to be 4.8s is 84.06m

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