JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Kinematics

Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at

t = 0

s. Ball B is thrown vertically down with an initial velocity '

u

' at

t = 2

s. After a certain time, both balls meet 100 m above the ground. Find the value of '

u

' in

ms^{- 1}

. [use

g = 10 ms^{- 2}

(A)10
(B)15
(C)20
(D)30✓ Correct

Answer: (D)

Solution

Ball A falls 80 m (from 180 m to 100 m) under free fall. Using

h = \frac{1}{2} g t^{2}

, the time taken is

t = \frac{2 \times 80}{10} = 4

s. Ball B is released 2 s later, so it travels for 2 s to meet A. For ball B, using

s = u t + \frac{1}{2} a t^{2}

with

s = - 80

m (downward),

t = 2

s, and

a = - 10

m/s²:

- 80 = - u (2) + \frac{1}{2} (- 10) (2)^{2}

. Solving:

- 80 = - 2 u - 20

, so

2 u = 60

and

u = 30

m/s.

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option

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