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2.Motion in Straight Line
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A ball is dropped from a building of height $45 \,m$. Simultaneously another ball is thrown up with a speed $40\, ms^{-1}$. Calculate the relative speed of the balls as a function of time.
Option A
Option B
Option C
Option D
Solution
For the ball dropped from the building, $u_{1}=0$, For the ball thrown up, $u_{2}=40 \mathrm{~m} / \mathrm{s}$. Velocity of the dropped ball after time $t$, $v_{1}=u_{1}+g t$ $v_{1}=g t$ (downward) For the ball thrown up, $u_{2}=40 \mathrm{~m} / \mathrm{s}$
Velocity of the ball after time $t$
$v_{2}=u_{2}-g t=(40-g t)$ (upward) $\ldots$ (2)
Relative velocity of one ball w.r.t. another ball
$=v_{1}-v_{2}$ $=g t-[-(40-\mathrm{gt})](\because$ From $(1)$ and $(2))$ $\quad=40 \mathrm{~m} / \mathrm{s}$
Standard 11
Physics
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