Answer: 4h
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
[tex]\Rightarrow v^2-u^2=2as\\[/tex]
Put [tex]v=0,a=-g[/tex] in above equation
[tex]\Rightarrow 0-u^2=2(-g)h\\\\\Rightarrow h=\dfrac{u^2}{2g}[/tex]
If the velocity of projection is doubled i.e. 2u
[tex]\Rightarrow h'=\dfrac{(2u)^2}{2g}\\\Rightarrow h'=4\times \dfrac{u^2}{2g}\\\\\Rightarrow h'=4h[/tex]
The maximum height becomes 4h
