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A ball is dropped from the top of a building $100 \mathrm{~m}$ high. At the same instant another ball is thrown upwards with a velocity of $40 \mathrm{~m} / \mathrm{s}$ from the bottom of the building. The two balls will meet after
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Verified Answer
The correct answer is:
$2.5 \mathrm{~s}$
Let balls meet after $t \mathrm{~s}$. The distance travelled by the ball coming down is
$s_1=\frac{1}{2} g t^2$
Distance travelled by the other ball
$\begin{array}{ll} & s_2=40 t-\frac{1}{2} g t^2 \\ \because \quad & s_1+s_2=100 \mathrm{~m} \\ \therefore \quad & \frac{1}{2} g t^2+40 t-\frac{1}{2} g t^2=100 \mathrm{~m} \\ & t=\frac{100}{40}=2.5 \mathrm{~s}\end{array}$
$s_1=\frac{1}{2} g t^2$
Distance travelled by the other ball
$\begin{array}{ll} & s_2=40 t-\frac{1}{2} g t^2 \\ \because \quad & s_1+s_2=100 \mathrm{~m} \\ \therefore \quad & \frac{1}{2} g t^2+40 t-\frac{1}{2} g t^2=100 \mathrm{~m} \\ & t=\frac{100}{40}=2.5 \mathrm{~s}\end{array}$
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