A ball is kiched at an angle ` 30^(@)` with the verical. If the horizontal componet of its velocity is `20 ms^(-1)`, find the maximum hight and hrizontal range. Use `= 10 ms^(-2)`.

Explanation

Here, theta =90^(@)-30^(@) =60^(@)`
Horizontal velocity, ` u cos 60^(@) =20 ms^(-1)`
or `u =(20)/( cos 60^(@)) =(20)/(1//2) =40 ms^(-1)`
Maximum height ,
`H=(u^(2) sin^(2) theta ) /(2 g) =(40^(2) sin ^260^(@))/(2 xx 10)`
`=(1600)/(2 g) =(40^(2) sin ^(2) 60^(@))/(10)`
`=(17600)/(10) (sqrt 3//2) =238.6 m`