Two balls $A$ and $B$ are placed at the top of $180 \mathrm{~m}$ tall tower. Ball $A$ is released from the top at $t=0 \mathrm{~s}$. Ball $B$ is thrown vertically down with an initial velocity $u$ at $t=2 \mathrm{~s}$. After a certain time, both balls meet $100 \mathrm{~m}$ above the ground. Find the value of $u$ in $\mathrm{ms}^{-1}$ [use $g=10 \mathrm{~ms}^{-2}$ ]

JEE PHYSICS MCQ: MOTION IN A STRAIGHT LINE

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Two balls

A

and

B

are placed at the top of

180 \mathrm{~m}

tall tower. Ball

A

is released from the top at

t=0 \mathrm{~s}

. Ball

B

is thrown vertically down with an initial velocity

u

t=2 \mathrm{~s}

. After a certain time, both balls meet

100 \mathrm{~m}

above the ground. Find the value of

u

\mathrm{ms}^{-1}

[use

g=10 \mathrm{~ms}^{-2}

]

View answer and explanation

Correct option: D

Option D (30) is correct because it is calculated by equating the distances traveled by both balls and solving for the initial velocity

u

of ball B.

Practice All Chapter MCQs →

\mathrm{g}

\mathrm{h}

\mathrm{u}

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

4.9 \mathrm{~m}

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

t

(t+1)

120 \mathrm{~m}

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

\mathrm{m}

(t+2)