A ball is thrown vertically upwards with a velocity of `49 m//s`. Calulate
(i) The maximum height to which it rises,
(ii) the total time it takes to return to the surface of the earth.

Explanation

Here, `u=49m//s , h=?`
total time of ascent + time of descent `=t+t=2t=?`
At maximum height, finaly velocity,`v=0`
acceleration `=-g=-9.8m//s^(2)`
from `v^(2)-u^(2)gh, 0-(49)^(2)=2(-9.8)h or h=(49xx49)/(2xx9.8)=122.5m`
from `v=u+g t, 0=49-9.8t or t=49/9.8=5s`
Total time`=2t,=10s`.