A tennis ball is thrown up and reaches a height of 4.05 m before coming down. What was its initial velocity? How much total time will it take to come down? Assume `g=10m//s^(2)`

Explanation

Given: Displacement (s) `=4.05m`
`g=-10m//s^(2)`
Final velocity `(v)=0m//s`
To find:
Initial velocity `(u)=?`
`t_("total")=?`
Formulae:
`v^(2)=u^(2)+2gs,=u+"gt`
`t_("total")=2xxt`
`:.v,g` and s are given
to find u.
We use 2nd equation
Solution:
a. `v^(2)=u^(2)+2gs`
`0^(2)=u^(2)+2xx(-10)xx4.05`
`0=u^(2)-81`
`u^(2)=81`
`u=9m//s` ........i
`:.` We have v, u and g to find t.
We use 1st equation
b. `v=u+gt`
`0=9+(-10)xxt` (from i)
`10t=9`
`t=0.9s`
Time of ascent `=` Time of descent
`t_("total")=2t`
`=2xx0.9`
`=1.8s`
The initial velocity was 9m/s and total time taken to come down is 1.8s.