Question

A BALL PROJECTED VERTICALLY UPWARDS FROM A, THE TOP OF TOWER

REACHES THE GROUND IN T1 SECOND. IF IT IS PROJECTED VERTICALLY

DOWNWARDS FROM A WITH THE SAME VELOCITY, IT REACHES THE GROUND IN T2

SECONDS. IF IT FALLS FREELY FROM A, SHOW THAT IT WOULD REACH THE GROUND

IN (T1T2)^1/2 S

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Answer 1

Answer:

h = -ua + (ga2/2) ..... (1) [using newton's second equation]

= = ub + (gb2/2)...... (2)

Gives: u = g(a-b)/2

putting back in (1), h = g(a2 + b - a) / 2

Now, Without any initial velocity, let the time taken to reach the ground be t.

then, h = gt2/2

t = √(2h/g) = √(a2 + b - a)

Explanation:

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