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Motion in a Plane

Section: Physics

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Question : 8 of 28

Marks: +1, -0

If a ball projected vertically upwards with certain intitial velocity from the ground crosses a point at a height of 25 m twice in a time interval of 4 s , then the initial velocity of the ball is

(Acceleration due to gravity

=10\, \mathrm{m}\, \mathrm{s}^{-2}

[AP EAPCET 21 May 2025 S2]

$20\, \mathrm{m}\, \mathrm{s}^{-1}$
$30\, \mathrm{m}\, \mathrm{s}^{-1}$
$40\, \mathrm{m}\, \mathrm{s}^{-1}$
$25\, \mathrm{m}\, \mathrm{s}^{-1}$

Solution:

If

v_1

be the velocity at the time of projection and

v_2

be the velocity at height 25 m , then

v_2^2 = v_1^2 - 2g \times 25

If

t

be the time taken to reach at maximum height

H

from point

B

, then

\;0 = v_2 - g t

\; v_2 = g t = 10 \times 2 = 20\, \mathrm{m/s}

\therefore

From eq. (i),

\; 20^2 = v_1^2 - 2 \times 10 \times 25

\Rightarrow 400 + 500 = v_1^2

\Rightarrow v_1 = \sqrt{900} = 30\, \mathrm{m/s}

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