A ball is thrown vertically upwards with a velocity of` 49 ms^-1`. Calculate :The total time it takes to return to the surface of earth.

Explanation

Initial velocity, `u = 48 ms^(-1)`
Final velocity, `V = 0 ms^(-1)`
Acceleration due to gravity,
`g = -9.8 ms^(-2)`
Distance = Height , h = ?
BY third law of motion,
`V^(2) - u^(2) = 2g s`
`rArr (0ms^(-1))^(2) - (48 ms^(-1))^(2) = 2 xx (-9.8 ms^(-2)) xx s`
`rArr -(48 ms^(-1))^(2) = 19.6 ms^(-2) xx s`
`rArr -2304 m^(2)s^(-2) = 19.6 ms^(-2) xx s `
`rArr (-2304 m^(2)s^(-2))/(-19.6 ms^(-2)) = s`
`rArr s = 117.55 m`
By first law of motion,
`V = u + g t`
`rArr (0 ms^(-1)) = 48 ms^(-1) + (-19.8 ms^(-2) xx t)`
`rArr -48 ms^(-1) = -19.8 ms^(-2) xx t`
`rArr (-48 ms^(-1))/(-9.8 ms^(-2)) = t`
`rArr 4.9 s = t`
`therefore` The maximum height to which the ball rises in `117.55`m.
The total time it takes to return to the surface of the Earth is
`4.9 s + 4.9 s = 9.8 s`.