A body is projected with a velocity `20 m*s^(-1)` , making an angle of `45^@` with the horizontal. Calculate
(i) the time taken to reach the ground `[g=10 m*s^(-2)]`,

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)`.
Let the total time of flight of the body be t. considering the vertical motion of the body , we get from the equation `h=ut-1/2"gt"^2`,
`0=10sqrt(2)t -1/2 *10*t^2` [as h=0]
`or, 5t^2 =10 sqrt(2)t`
`therefore` Total time of flight,
`t=(10sqrt(2))/(5)=2sqrt(2) =2xx1.414=2.828` s.