a ball is thrown with `25 m//s` at an angle `53^(@)` above the horizontal. Find its time to fight, maximum height and range.

Explanation

In the adjining figure velocity of projection `u=25 m//s`, angle of projection `theta=53^(@)`, the horizontal and vertical components `u_(x)` and `u_(y)` of velocity of projection are shown. From these information we have
`u_(x)=u cos 53^(@)=15 m//s` and `u_(y)=u sin 53^(@)=20 m//s`
Using equations for time of flight T, maximum height H and range R,
we have `T=(2u_(y))/g=(2xx20)/10=4 s`
`H=u_(y)^(2)/(2g)=20^(2)/(2xx10)=20 m`
`R=(2u_(x)u_(y))/g=(2xx15xx20)/10=60 m`