NCERT

Class 9

Science

Solutions

A ball thrown up vertically returns to the thrower after 6s find.

(a) the velocity with which it was thrown up.

(b) the maximum height it reaches and

(c) it's position after 4s.

Time taken to reach the maximum height

t = 6/2

t = 3s

v = 0

(a) u = ?

from the equation, v = u – gt

0 = u - 9.8 x 3

-u = -29.4

u = 29.4 m/s

(b) from the equation. v^{2} = u^{2} – 2gh

0 = (29.4)^{2} – 2 x 9.8 x h

19.6h = 29.4 x 29.4

h = (29.4 x 29.4) / 19.6

h = 44.1 m

Thus the maximum height reached by the ball, 44.1 m

(c) t = 4s (given)

but the ball reaches the maximum height after 3s. Therefore the distance covered after falling from the maximum height after 1s so t = 1s.

h = ut + 0.5 gt^{2}

h = ( 0 x 1) + (0.5 x 9.8 x 1 x 1)

h = 0 + 4.9

h = 4.9 ~ 5m

∴ Height above the ground, H = 44.1 - 4.9

H = 39.2 ~ 39m

Hence after 4 s, the ball is at a height of 39 m above the ground.

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