Question

A body thrown horizontally from the top of a building with a speed 20 m//s, strikes the ground 40 m away from the foot of tower. The height of the tower is

Answer 1

Answer: 20 m

Explanation:

In horizontal projectile motion, the time period is same as that of a freely falling body from the same height.

So,

t = √(2H/ g).............(1)

Now along the horizontal direction (i.e. x axis),

s = ut + 1/2at²

i.e., x = ut + 1/2(0)t² [Since there is no acceleration along horizontal direction, a = 0]

=> x = ut

=> x = u . √(2H/ g) (From 1)

Then on rearranging,

H = x²g/2u² = R²g/2u²

Here , 'x' , which the horizontal distance covered, is same as that of the range.

On substitution,

H = 1600 x 10/2 x 400

= 20 m

Hope it helps.

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