A ball is thrown upwards from the ground with an initial speed of u. The ball is at height of `80m` at two times, the time interval being 6 s. Find u. Take `g=10 m//s^2.`

Explanation

Here, `a=g=-10 ms^(-2)` and `s=80 m`
Substituting the values , in `s=ut+(1)/(2)at^(2)`,
We have `80 = ut-5t^(2)`

or `5t^(2)-ut^(2)+80=0`
or `5t^(2)-ut+80=0`
or `t=(u+sqrt(u^(2)-1600))/(10)` and `t=(u-sqrt(u^(2)-1600))/(10)`
Now, it is given that.
`(u+sqrt(u^(2)-1600))/(10)-(u-sqrt(u^(2)-1600))/(10)=6 rArr (sqrt(u^(2)-1600))/(5)=6`
`rArr sqrt(u^(2)-1600)=30rArr u^(2)-1600=900`
`therefore u^(2)=2500 rArr u= +- 50 ms^(-1)`
Ignoring the negative sign, we have
`u=50 ms^(-1)`