A stone is thrown vertically upwards with a velocity of `20ms^(-1)` from the top of a tower 96 m high. After what time will it reach the foot of the tower ? What will be the velocity of the stone reaching the point of projection? At what velocity will it strike the ground ? `(g=9.8 ms^(-2))`

Explanation

Given `v_(0)=20ms^(-1), g=9.8 ms^(-2)`.
`"using h = 96 m"`
`h=v_(0)t+1//2g t^(2)` where
`h=-h, g=-g ` we get
`-96=20t -t//2 xx9.8t^(2)`
`-4.9 t^(2)-20t+96=0`
`"or "4.9 t^(2)-20t-96=0`
`therefore" "t=(20pm sqrt(480+1881.6))/(2xx4.9)=(20pm 47.77)/(9.8)`

`"i.e. "t=(67.77)/(9.8)=6.91 s`
Velocity of the stone reaching the point of projection `=-20ms^(-1)`
By applying the formula `v^(2)=v_(0)^(2)+2ax`
we get `v^(2)=v_(0)^(2)+2(-g)(-h)=(20)^(2)+2xx9.8 xx96=2281.6`
`therefore" "v=47.77ms^(-1)`