A ball is projected vertically upward with an initial velocity of $50 \mathrm{~ms}^{-1}$ at $\mathrm{t}=0 \mathrm{~s}$. At $\mathrm{t}=2 \mathrm{~s}$, another ball is projected vertically upward with same velocity. At $\mathrm{t}=$ _____ s, second ball will meet the first ball $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$.

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Kinematics

A ball is projected vertically upward with an initial velocity of

50 ms^{- 1}

t = 0 s

. At

t = 2 s

, another ball is projected vertically upward with same velocity.

At

t =

_____ s, second ball will meet the first ball

(g = 10 ms^{- 2})

Answer: 6

Solution

Let the balls meet at time

t

. The first ball travels for time

t

and the second for time

(t - 2)

. They meet at the same height, so:

50 t - \frac{1}{2} g t^{2} = 50 (t - 2) - \frac{1}{2} g (t - 2)^{2}

. Simplifying:

100 = \frac{g}{2} [t^{2} - (t - 2)^{2}] = \frac{10}{2} [4 t - 4]

, which gives

20 = 4 t - 4

, so

t = 6

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Integer answer

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