Q. 18. A ball thrown up vertically returns to the thrower after 6 s. Find
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s.
(a) Let the ball be thrown with initial velocity u
Time taken to get back to the thrower = 6 s
Time taken to reach the highest point is t = 6/2 = 3 s
Final velocity at the highest point is v = 0
Acceleration a = -g = -9.8 m s-2
Using the first equation of motion
(b) Let the maximum height it reaches be s
Using the second equation of motion
(c) Out of the 4 seconds, 3 have been spent in reaching the highest point
The distance travelled by the ball in the next 1 second is s' given by
Distance from the ground after 4 seconds = s - s' = 44.1 - 4.9 = 39.2 m
The position of the ball after 4 seconds is 39.2 m from the ground.