Question

A body is projected with a velocity ` 30 ms^(-1)` at an angle of `30^@` with the vertical. Find (i) the maximum height (ii) time of flight and (iii) the horizontal range of the projectile. Take ` g=10 m//s^@`.

Answer 1

Here, ` u= 30 ms^(-1)` , Angle of projection with horizontal, ` theta = 90^@ = 60^@.
(i) Max. height, ` H= (u^2 sin^2 theta)/(2 g) = ((30)^2 sin^2 60^@)/ ( 2xx 10) = 33 . 75 m`
(ii) Time of flight,
`T = ( 2 u sin theta )/ g = ( 2 xx 30 2 xx 60^2)/(10) = 5.2 s`
(iii) Horizontal range,
` R= (u^2 sin theta0/g = ((30^sin 2 xx 60^2) /(10)`
`= ( 900) /(10) sin 120^@ = 90 sin 60^@ = 77 . 94 m`.