Question

From the top of a tower T, a ball is thrown vertically upwards. When the ball reaches a distance 'h' below T, its speed is double of want is was at height 'h' above T. Show that the greatest height attained by the ball above T is 5h/3.

Answer 1

Let AT = (x + h) be the maximum height attained by the ball.

Let B, the point at a height 'h' above T. Let at B, the velocity of the ball be 'v₁' Therefore velocity attained by the ball to reach B from A can be obtained using.

v² - u² = 2as

or, v₁² - 0² = 2gx

or, v₁ = √2gx.........(i)

Let v₂ be the velocity of the ball at C whose distance from 'T' is 'h'.

Squaring both sides, we get

2g(2h + x) = 4 x 2gx

or, 2h + x = 4x

or, 3x = 2h

or, x = 2h/3

Maximum height attained by the ball

= AT = h + x = h + 2/3h = 5/3h.