A ball is thrown vertically upwards with a velocity of `50ms^(-1)` and the same later thrown at an angle for the maximum range. Compare the heights reached in the two cases `(g=9.8ms^(-2))`

Explanation

`u=50ms^(-1),g=9.8ms^(-2)`
(i) For a vertical throw,
`v^2=v_0^2-2gh` here v=0, `:.h=(v_0^2)/(2g)`
`h=(50xx50)/(2xx9.8)=127.5m`

For a projectile motion `R_("max")=(v_0^2)/(g)" and "R_("max")=4H`
`:.H=(R_("max"))/4=(v_0^2)/(2g)xx1/4=(127.5)/4=31.875m`
`h: H: :4:1.