A particle is thrown vertically upwards from the top of a tower with a certain velocity and takes 8s to reach the foot of the tower. The same when thrown with the same speed downwards, it reaches the foot of the tower in 2s. Calculate what time the particle would take to reach the foot of the tower when dropped freely. Taking `g=9.8ms^(-2)` calculate the height of the tower and the initial speed of the projectile.

Explanation

We know that,
`t=sqrt(t_(1)t_(2))=sqrt(8xx2)" "t=4s`
Using `h=1//2g t^(2)`
We get height of the tower `h=1//2xx9.8xx4^(2)=78.4m`
Using the formula `h=v_(0)t+1//2 g t^(2)`, where
taking `h=-h, g = -g`, for a vertical projection
and `t=t_(1)=8s` we get,
`-78.4=v_(0)(8)-1//2xx9.8xx8^(2)`
`"i.e. "-78.4+313.6=8u`
`therefore" "v_(0)=(235.2)/(8)=29.4ms^(-1)`
Initial velocity of the particle is `29.4ms^(-1)`.