Question

A ball is dropped from the top of a tower. It acquires a velocity 20 m/s on reaching the ground. Calculate the height of the tower. g=10m/s^2.
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Answer 1

Explanation:

let height of the tower=h

initial velocity (u)=0

final velocity (v)=20m/s

acceleration due to gravity (a)=10m/s^2

we know kinematic equation: v^2-u^2=2as

where, s=h=height of tower

substitute the given values:

(20)^2-(0)^2=2(10)(h)

400=20h

so, h=400/20=20

so, height of tower=20 metres

hope it helps you!

Answer 2

Answer:

Since the ball is dropped from the top of a tower,

Initial velocity (u)= 0 m/s

Final veocity(v)= 20 m/s

Acceleration due to gravity = g= 10 m/s²

By third equation of motion,

v²=u²+2as

Or, 20²=0²+2*10*s

Or, 20s= 400

Or, s= 400/20= 20

Hence ,the height of the tower is 20 m.