A body is projected with a velocity `20 m*s^(-1)` , making an angle of `45^@` with the horizontal. Calculate
(ii) horizontal range.

Explanation

The vertical and horizontal components of the velocity of `20m*s^(-1)` are
`u_(H) =20 cos 45^@ =20xx1/(sqrt(2)) =10sqrt(2) m*s^(-1)`
and `u_(V)=20 sin 45^@ =20 xx1/sqrt(2)=10sqrt(2) m*s^(-1)`.
(iii) Let the distnace from the point of projection to the point at the ground where the body touches be x .
By considering the horizontal motion of the body , we get,
`x=u_(H)xxt =10sqrt(2)xx2sqrt(2)=40m`.
`therefore` The horizontal range =40 m.