JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Kinematics

A ball is thrown vertically upwards with a velocity of

19.6 ms^{- 1}

from the top of a tower. The ball strikes the ground after 6 s. The height from the ground up to which the ball can rise will be

(\frac{k}{5}) m

. The value of k is ..... (use

g = 9.8 m / s^{2}

)

Answer: 392

Solution

Time to reach maximum height from the tower top:

t_{a} = \frac{u}{g} = \frac{19.6}{9.8} = 2

s.

Time to fall from maximum height to ground:

t_{d} = 6 - 2 = 4

s.

Maximum height from ground:

h_{max} = \frac{1}{2} g t_{d}^{2} = \frac{1}{2} \times 9.8 \times 16 = 78.4

m.

Alternatively, using the given form:

h_{max} = \frac{k}{5}

, and from the solution

h_{max} = \frac{392}{5}

, we get

k = 392

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Integer answer

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