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.

Answers (1)
S Sayak

(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

\\v=u+at\\ 0=u-9.8\times 3\\ u=29.4\ m\ s^{-1}

(b) Let the maximum height it reaches be s

Using the second equation of motion

\\s=ut+\frac{1}{2}at^{2}\\ s=29.4\times 3-4.9\times 3^{2}\\ s=44.1m

(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

\\s'=ut+\frac{1}{2}gt^{2}\\ s=0\times +4.9\times 1^{2}\\ s=4.9m

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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