Question

a ball is released from the top of a tower of height h meter .it takes T second to reach ground .what is the position of ball above the ground T /5 second​

Answer 1

Initial velocity of ball, u = 0

given, height of tower is h m .

time taken by ball to reach the ground is T.

using formula

s = ut + 1/2at²

here , s = -h , u = 0, a = -g and t = T

then, -h = 0 - 1/2gT²

or, T² = 2h/g ......(1)

now, at t = T/5 position of ball is s (let)

using formula,

s = ut + 1/2 at²

or, -s = 0 + 1/2 (-g)(T/5)²

or, s = gT²/50

from equation (1),

s = g(2h/g)/50 = h/25 m

hence, position of particle from the ground = h - h/25 = 24h/25