A ball is thrown vertically upwards with a velocity of 49 ms -- Calculate fi) the maximum height to which it rises, (ii) the total time it takes to return to the surface of the earth.

Explanation

As per Cartesian sign convention, upward velocity is taken as positive and the acceleration due to gravity (acting in the downward direction) is taken as negative.
`u=+49ms^(-1), g=-9.8ms^(-2)`
`v^(2)-u^(2)=2gs`
`0-(49)^(2)=2(-9.8) xx h (therefore s=h)`
`h=(49 xx 49)/(2 xx 9.8)=122.5m`
(ii) Let t be the time taken by the ball to reach the highest point.
v=u+gt
0=49+(-9.8)t
`t=(49)/(9.8)`
t=5 sec
Now, time of ascent=time of descent
Total time taken to return to the surface of the earth `=2 xx t=2 xx =10sec`