Assume that a ball is kicked at an angle of `60^(@)` with the horizontal, so if the horizontal component of its velocity is `19.6 ms^(-1)`, determine its maximum height.

Options

A. `58.8 m`
B. `40 m`
C. `120 m`
D. `60 m`

Answer

A

Explanation

Given, `theta = 60^(@)`
Horizontal component of velocity `= u cos 60^(@) = 19.6 ms^(-1)`
`:. U = (19.6)/(cos 60^(@)) = (19.6)/(0.5) = 39.2 ms^(-1)`
Therefore, maximum height,
`H = (u^(2) sin^(2) 60^(@))/(2g) = ((39.2)^(2))/(2 xx 9.8) xx ((sqrt(3))/(2))^(2) = 58.8m`