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 = 30^@, u cos theta = 20 ms^(-1)`
or `u cos 30^@ =20`
or ` u = (20) /( cos 30^@) = ( 20) / ( sqrt3/2) = ( 40)/( sqrt 3) ms^(-1)`
Maximum height
` H= (u^2 sin^2 theta)/( 2g) = (( 40// sqrt 3)^2 sin ^2 30^@0/( 10) = 6.67 m`
Horizontal range
`= (u^2 sin 2 hteta )/g = (( 40//sqrt3)^2 sin 2 xx 30^2)/ ( 10) = 46.19 m`.