JEE PHYSICS MCQ: MOTION IN A STRAIGHT LINE

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A stone thrown vertically upwards with a speed of

5 \mathrm{~m} / \mathrm{sec}

attains a height

H_1

. Another stone thrown upwards from the same point with speed of

10 \mathrm{~m} / \mathrm{sec}

attains a height

H_2

. The correct relation between

H_1

and

H_2

is:

View answer and explanation

Correct option: A

{H}_2=4 H_1

Option A (

{H}_2=4 H_1

) is correct because the maximum height is proportional to the square of the initial velocity, and the second stone's initial velocity is twice that of the first.

More Questions from MOTION IN A STRAIGHT LINE

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Assertion: In a free fall, the initial velocity of a body may not be zero.
Reason: Free fall means the vertical acceleration of the body is equal to

\mathrm{g}

. A ball is thrown vertically upward with speed 15 m/s and it returns to the ground with speed 10 m/s as a constant air resistance acts. The maximum height attained by the ball is Which of the following is not a derived quantites? Assertion: If a body is dropped from the top of a tower of height

\mathrm{h}

and another body is thrown up simultaneously with velocity

\mathrm{u}

from the foot of the tower, then both of them would meet after a time

\frac{\mathrm{h}}{\mathrm{u}}

.
Reason: For a body projected upwards, the distance covered by the body in the last second of its upward journey is always

4.9 \mathrm{~m}

irrespective of velocity of projection.

(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2)

A particle is having uniform acceleration. If its displacement from

t

(t+1)

second is

120 \mathrm{~m}

and change in velocity is

50 \mathrm{~m} / \mathrm{s}

. Find its displacement (in

\mathrm{m}

) in

(t+2)

second. An aeroplane flies 400m north and 300m south and then flies 1200m upwards, then net displacement is