Question

A stone is thrown vertically upwards from the ground with a velocity 15 m/s. At the same instant a ball is dropped from a point directly above the stone from a height of 30 m. At what height from the ground will the stone and the ball meet ?
(Use g=10m/s2g=10m/s2 for ease of calculation).
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Answer 1

let a stone and a ball meet at height h from the ground and time taken to meet the stone and the ball is t.

A stone is thrown vertically upwards from the ground with velocity 15 m/s.

using formula, s = ut + 1/2 at²

here, s = h, u = 15m/s , a = -g = -10 m/s²

⇒ h = 15t - 1/2 (10)t²

⇒h = 15t - 5t² ..........(1)

at the same time , a ball is dropped from 30m height.

so, u = 0, s = -(30 - h), a = -g

so, -(30 - h) = 0 - 1/2(10)t²

⇒-30 + h = - 5t²

from equations (1),

-30 + 15t - 5t² = -5t²

⇒15t = 30

⇒t = 2

now height from the ground, h = 15(2) - 5(2)² = 30 - 20 = 10m

Answer 2

$\huge\bold\purple{Answer:-}$

let a stone and a ball meet at height h from the ground and time taken to meet the stone and the ball is t.

A stone is thrown vertically upwards from the ground with velocity 15 m/s.

using formula, s = ut + 1/2 at²

here, s = h, u = 15m/s , a = -g = -10 m/s²

⇒ h = 15t - 1/2 (10)t²

⇒h = 15t - 5t² ..........(1)

at the same time , a ball is dropped from 30m height.

so, u = 0, s = -(30 - h), a = -g

so, -(30 - h) = 0 - 1/2(10)t²

⇒-30 + h = - 5t²

from equations (1),

-30 + 15t - 5t² = -5t²

⇒15t = 30

⇒t = 2

now height from the ground, h = 15(2) - 5(2)² = 30 - 20 = 10m