A football is kicked at an angle of `30^(@)` with the vertical, so if the horizontal component of its velocity is `20 ms^(-1)`, determine its maximum height.

Explanation

Given, `theta = 30^(@)`
Horizontal component of velocity `= u sin 30^(@) = 20 ms^(-1)`
`u = (20)/(sin 30^(@)) = (20)/(1//2) = 40 ms^(-1)`
Therefore, maximum height,
`H = (u^(2) cos^(2) 30^(@))/(2g) = ((40)^(2))/(2 xx 9.8) xx ((sqrt(3))/(2))^(2) = 61.22 m`