A body is projected with a velocity `20 m*s^(-1)` , making an angle of `45^@` with the horizontal. Calculate
the maximum height it can attain

Explanation

The vertical and horizontal components of the velocity of `20m*s^(-1)` are
`u_(H) =20 cos 45^@ =20xx1/(sqrt(2)) =10sqrt(2) m*s^(-1)`
and `u_(V)=20 sin 45^@ =20 xx1/sqrt(2)=10sqrt(2) m*s^(-1)`.
(ii) Let the maximum height attained be h . Vertical velocity at the maximum height =0.
Considering the vertical motion of the body , we get from the equation `v^2=u^2-2gh`,
`0= (10sqrt(2))^2-2*10*h therefore h=10m `.