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2.Motion in Straight Line
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A ball is dropped from the top of a building $100\,m$ high. At the same instant another ball is thrown upwards with a velocity of $40\,m / s$ from the bottom of the building. The two balls will meet after $..........\,s$
A
$3$
B
$2$
C
$2.5$
D
$5$
(AIIMS-2016)
Solution
(c)
Let balls meet after $t s$. The distance travelled by the ball coming down is
$s_1=\frac{1}{2} g t^2$
Distance travelled by the other ball
$s_2=40 t-\frac{1}{2} g t^2$
$\because \quad s_1+s_2=100\,m$
$\therefore \quad \frac{1}{2} g t^2+40 t-\frac{1}{2}\,g t^2=100\,m$
$t=\frac{100}{40}=2.5\,s$
Standard 11
Physics
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