Question

A body is thrown up with with velocity u.it reaches maximum height h. If velocity of projection is doubled the maximum height is..

Answer 1

Answer: 4h

Step-by-step explanation:

Given

The body is thrown vertically upward with velocity u and reaches a height of h

The relation between u and h is given by

`\Rightarrow v^2-u^2=2as\\`

Put
`v=0,a=-g` in above equation

`\Rightarrow 0-u^2=2(-g)h\\\\\Rightarrow h=(u^2)/(2g)`

If the velocity of projection is doubled i.e. 2u

`\Rightarrow h'=((2u)^2)/(2g)\\\Rightarrow h'=4* (u^2)/(2g)\\\\\Rightarrow h'=4h`

The maximum height becomes 4h