Question

3. A stone is dropped from the roof of a tower of height h. The total distance covered by the stone in the last 2
seconds of its motion is equal to the distance covered by it in the first four seconds. Find the height of the
tower.
​

Answer 1

Answer:

h = 125 m

Explanation:

Hello Friend,

Lets say that the stone is in motion for ' t ' seconds.

Distance covered by the stone in first 4 seconds:

u = 0 , t =4 , a = 10

s = 1/2 x 10 x 16 = 80 m

In last two seconds:

The stone would already have some velocity due to it's drop from intitial position. Let us find this velocity:

t = t -2 (since only two seconds are left in motion) , u = 0 , a =10

v = 10(t-2)

The stone would have a velocity of 10 (t-2) two seconds before it hits the ground.

So,

u = 10(t-2) , a = 10 , t =2 (as it is given only two seconds are left in motion) ,         s = 80 (as it is given distance covered in first 4 seconds is equal to this distance)

80 = 10(t-2)2 + 5(4) => 80 = 20t - 40 + 20 => t = 5 seconds

Therefore the stone is in motion for 5 seconds.

IN the questions, they demand us to find h.

h = 1/2 x 10 x (25)

Therefore h = 125 m

Dear friends, I have directly applied equations of motions here. Please understand it, I used only s = ut + 1/2 at^2 and v = u + at , nothing else.

If you like the answer please mark me as brainliest and also give me thanks.

Answer 2

Answer:

Height of the tower is 361.25m

Explanation:

Explanation is in the attachment..

Hope it will help!