The distance, s meters, which a falling body covers in time t seconds, is 4.8 times the square of the time t.
A stone which is dropped from a cliff top takes 35 seconds to reach the base of the cliff. Calculate the height of the cliff to the nearest meter.

Open in App

Solution

Let's form the equation using given information,
$s = 4.8 t^{2}$
Given $t = 35$ seconds, Let's substitute value of $^{'} t^{'}$ in above equation
$∴ s = 4.8 \times 35^{2} = 5880 m$

Suggest Corrections

Similar questions

Q. The distance s metres, which a falling body covers in time t seconds is 4.9 times the square of the time t. A stone which is dropped from a cliff top takes 3.5 seconds to reach the base of the cliff. Use this information to calculate the height of the cliff to the nearest metre.

Q. A stone is kicked horizontally from the edge of a cliff. The stone is kicked at

7 m / s

and takes

3

seconds to reach the ground at the base of the cliff.
Which one of the following is closest to the height of the cliff?

From the top of a cliff of $92 m$ height, the angle of depression of a buoy is $20^{\circ}$ . Calculate to the nearest meter, the distance of the buoy from the foot of the cliff. {Use $tan 20^{o} = 0.36$ }

Q. A stone falls from a cliff and travels

34.3 m

in the last second before it reaches the ground. Calculate the height of the cliff.

Q. A stone is dropped from the top of a tall cliff and n seconds later another stone is thrown vertically downwards with velocity u. Then the second stone overtakes the first, below the top of the cliff at a distance given by:

[Assume u sufficiently enough to overtake the first stone]

Join BYJU'S Learning Program

The distance, s meters, which a falling body covers in time t seconds, is 4.8 times the square of the time t.A stone which is dropped from a cliff top takes 35 seconds to reach the base of the cliff. Calculate the height of the cliff to the nearest meter.

Let's form the equation using given information,s=4.8t2Given t=35 seconds, Let's substitute value of ′t′ in above equation ∴s=4.8×352=5880m

The distance, s meters, which a falling body covers in time t seconds, is 4.8 times the square of the time t.
A stone which is dropped from a cliff top takes 35 seconds to reach the base of the cliff. Calculate the height of the cliff to the nearest meter.

Let's form the equation using given information,
$s = 4.8 t^{2}$
Given $t = 35$ seconds, Let's substitute value of $^{'} t^{'}$ in above equation
$∴ s = 4.8 \times 35^{2} = 5880 m$