JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Kinematics

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From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in ______ s.

Answer: 3

Solution

Let the height of the tower be

h

and the initial speed be

u

.

For the first ball (thrown upward, taking downward as positive):

h = - u (6) + \frac{1}{2} g (6)^{2} = - 6 u + 18 g ...(1)

For the second ball (thrown downward):

h = u (1.5) + \frac{1}{2} g (1.5)^{2} = 1.5 u + 1.125 g ...(2)

Multiplying equation (2) by 4:

4 h = 6 u + 4.5 g ...(3)

Adding equations (1) and (3):

5 h = 22.5 g

h = 4.5 g

For the third ball (released from rest):

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

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

t^{2} = 9

t = 3 s

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Integer answer

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JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Kinematics

Solution

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