Question

A stone is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stopwatch measures the stone's trajectory time from the top of the cliff to the bottom to be 4.3 s. What is the height of the cliff if air resistance is negligibly small?

Answer 1

Answer:

The height of the cliff is 90.60 meters.

Explanation:

It is given that,

Initial horizontal speed of the stone, u = 10 m/s

Initial vertical speed of the stone, u' = 0 (as there is no motion in vertical direction)

The time taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s

Let h is the height of the cliff. Using the second equation of motion in vertical direction to find it. It is given by :

$h=u't+\dfrac{1}{2}gt^2$

$h=\dfrac{1}{2}gt^2$

$h=\dfrac{1}{2}\times 9.8\times (4.3)^2$

h = 90.60 meters

So, the height of the cliff is 90.60 meters. Hence, this is the required solution.

Answer 2

The height of the cliff is 90.60 meters.

Calculation of the cliff height:

Since

The initial horizontal speed of the stone, u = 10 m/s

The initial vertical speed of the stone, u' = 0 because there is no motion in the vertical direction

The time is taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s

Here we assume h is the height of the cliff.

Now second equation of motion in vertical direction should be used.

$h = 1\div 2 \times 9.8 \times 4.3^2$

= 90.60 meters.

Simply we applied the above formula to determine the height.

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