An object thrown vertically upwards reaches a height of 500 m. What was its initial velocity? How long will the object take to come back to earth? Assume `g=10m//s^(2)`

Explanation

Given :
`g=10m//s^(2)andv=500m`
To find : Initial velocity (u),
Time taken (t)
Formula : (i) `s=ut+at^(2)`
(ii) v = u + at
(iii) `v^(2)=u^(2)+2as`
Solution :
At max height v = 0
`a=-g=-10m//s^(2)`
s = h
`v^(2)=u^(2)+2as`
`0=u^(2)-2(10)(500)`
`u^(2)=10000`
`u=100m//s`
v = u + at
0 = 100 - (10)t
10t = 100
t = 10 s
Total time = 10 + 10 = 20 s.
The object will take 20 seconds to come back to the earth.