A ball is thrown from a field with a speed of 12.0 m/s at an angle of `45^0` with the horizontal. At what distance will it hit the field again ? Take `g=10.0 m/s^2`

Explanation

The horizontal range `= (u^2sin 2 theta)/g`
`=((12 m/s)^2xxsin(2xx45^0))`
`=(144 m^2/s^2)/(10.0m/s^2) = 14.4 m.
Thus, the ball hits the field at 14.4 m from the point of projection.