MCQ
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
    $2$
  • $2.5$
  • D
    $5$

Answer

Correct option: C.
$2.5$
c
(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

$s_2=40 t-\frac{1}{2} g t^2$

$\because \quad s_1+s_2=100\,m$

$\therefore \quad \frac{1}{2} g t^2+40 t-\frac{1}{2}\,g t^2=100\,m$

$t=\frac{100}{40}=2.5\,s$

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