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.

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