A ball is dropped from a highly raised platform at t = 0 starting… - Vidyadip

MCQ

A ball is dropped from a highly raised platform at t = 0 starting from rest. After 6 second another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v? $\left(\right.$ Take $\left.g =10 m / s ^2\right)$

✓
$75 m / s$
B
$55 m / s$
C
$40 m / s$
D
$60 m / s$

✓

Answer

Correct option: A.

$75 m / s$

(A)
Let the two balls meet after t at distance x from the platfrom.
Using $h = ut +\frac{1}{2} gt ^2$
For the first ball
$u = 0 , t =18 s, g =10 m / s ^2$
$\therefore \quad x =\frac{1}{2} \times 10 \times(18)^2... (i)$
For the first secound ball
$u = v , t =12 s, g =10 m / s ^2$
$\therefore \quad x = v \times 12+\frac{1}{2} \times 10 \times(12)^2... (ii)$
From equations (i) and (ii),
$\frac{1}{2} \times 10 \times(18)^2=12 v +\frac{1}{2} \times 10 \times(12)^2$
$12 v =\frac{1}{2} \times 10 \times\left[(18)^2-(12)^2\right]=\frac{1}{2} \times 10 \times 180$
$v =\frac{1 \times 10 \times 30 \times 6}{2 \times 12}=75 m / s$

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