A body is projected at an angle of `30^@` with the horizontal and with a velocity of `39.2 ms^(-1)`. Find ,
A. Time of flight B. Range ( R) C. Maximum height ( H)

Explanation

Give `theta= 30^@` , u=39.2. `ms^(-1)`
Given time of fight , `T=(2u sin theta)/(g)= (2xx39.2 xx sin 30^(@))/(9.8)`
Range `R= (u^2 sin 2 theta)/(g) = (39.2 xx 39.2 xx sin (2xx 30^@))/(9.8)`
`therefore R= 135.79m`
Maximum height `H=(u^2 sin theta)/(2g) , sin 30^@= 1//2`
`= (39.2 xx 39.2 xx (1//2)^2)/(2xx 9.8)`
`therefore H= 19.6 m`