A stone thrown upwards with a speed u from the the top of the tower reaches the ground with a speed 3u. The height of the tower is

Options

A. `(3u^(2))/g`
B. `(4u^(2))/g`
C. `(6u^(2))/g`
D. `(9u^(2))/g`

Answer

B

Explanation

1 methode -Let downward direction is taken as +ve. Initial is -ve=u(say)
`:.` From the equation, `v^(2)-u^(2)=as` we get `(3u)^(2)-(-u)^(2)=20 hg`
`rArr h=(u^(2))/g` "B"
The stone is thrown vertically upward with an initial velocity u from the top a tower it reaches the highest point and returns back and reaches the tope of tower with the same velocity u vertically downwards. Now, from the equation
`rArr (3u)^(2)=u^(2)+2gh rArr 2gh =9u^(2)-u^(2) rArr h=(8u^(2))/(2g) rArr h=(4u^(2))/g` "B"