An object is projected with a velocity of `30 ms^(-1)` at an angle of `60^(@)` with the horizontal. Determine the horizontal range covered by the object.

Explanation

Given, initial velocity, `u = 30 ms^(-1)`
Angle of projection, `theta = 60^(@)`
Therefore, the horizontal range (or distance) covered by the object will be given as
`R = (u^(2) sin 2 theta)/(g) = ((30)^(2) sin (60^(@)))/(g)`
`=((30)^(2) sin 60^(@))/(9.8) = 79.53 m`
`rArr R = 79.53 m`