2. A body is thrown vertically upwards from the top of a tower of height 54.2m with an initial velocity of 21.6 m/s. Taking g as 10 m/s², calculate :-(i)the height to which it will rise before returning to the ground, (ii) the velocity with which it will strike the ground and (iii) the total time of the journey.
Answers
Explanation:
(i) Let the initial velocity be u, final velocity be v, distance travelled upwards before the velocity becomes zero be s.
We know,
v²=u² - 2gs (due to retardation)
0= 21.6² - 2×10×s (u=21.6 v= 0)
466.56÷20 = s
s = 23.328m
Total height = 54.2 + 23.328=77.528m
Therefore it will rise to a height of 77.528m above the ground.
(ii) v² = u²+ 2gs
Here v= final velocity with which it will strike the ground u = 0 s = 77.528
v²= 20×77.528
v = √1550.56
v = 39.377m/s
(iii) v = u - gt (due to retardation)
While travelling upward u = 21.6, v = 0,
Therefore,
21.6÷10 = t
2.16s = t
While travelling downward u = 0, v = 39.377
Therefore,
39.377÷10 = t
3.9377s = t
Total time of journey = 3.9377+2.16
=6.0977s