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Two balls $\mathrm{A}$ and $\mathrm{B}$ are placed at the top of $180 \mathrm{~m}$ tall tower. Ball A is released from the top at $\mathrm{t=0 \mathrm{~s}}$ . Ball $\mathrm{B}$ is thrown vertically down with an initial velocity $\mathrm{' u ' \: at\: t=2 \mathrm{~s}}$ . After a certain time, both balls meet $100 \mathrm{~m}$ above the ground. Find the value of $' \mathrm{u}$ ' in $\mathrm{ms}^{-1}$ . [use $g=10 \mathrm{~ms}^{-2}$ ] :

Option: 1
$10$

Option: 2
$15$

Option: 3
$20$

Option: 4
$30$

Answers (1)

best_answer

Let they meet $\mathrm{t=t_{\circ }}$

For A

$\mathrm{80=\frac{1}{2}\; g\; t{_{\circ }}^{2}}$ ........(i)

For B

$\mathrm{80=u\left(t_{0}-2\right)+\frac{1}{2} g\left(t_{0}-2\right)^{2}.......\text { (ii) }}$

From (i)

$\mathrm{t_{\circ }=4}$

$\mathrm{80=u \times 2+5 \times 2^{2}}$

$\mathrm{u=30\; m/s}$

Correct answer is Option (4)

Posted by

Ritika Harsh

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