Question

a stone is dropped from the top of a tower of height 40m.at same time stone b is projected from the bottomof tower with an initial velocity such that stone coolided midway .the intitial speed of second stone b will be

Answer 1

Answer:

• Initial velocity of second stone will be 14 √2  ms⁻¹.

Explanation:

Since, first stone is dropped from the top of tower therefore,

initial velocity of first stone, u₁ = 0

Let, initial velocity of second stone = u₂ = ?

height of tower is 40 m and stone colloid at midway therefore,

distance travelled by first stone before collision = distance travelled by second stone before collision = s = 20 m

Let, time after which stones collide = t

then,

Using second equation of motion

For first stone, taking acceleration due to gravity, g = 9.8 ms⁻¹

→ s = u₁ t + 1/2 g t²

→ 20 = (0) t₁ + 1/2 (9.8) t²

→ 20 = 4.9 t²

→ t² = 20/4.9

→ t = (10√2) / 7   sec

Now,

again Using second equation of motion

For second stone, taking acceleration due to gravity, g = -9.8 ms⁻¹

→ s = u₂ t + 1/2 g t²

→ 20 = u₂ ((10√2) / 7) + 1/2 (-9.8) ((10√2)/7)²

→ 20 = (10√2 u₂ / 7)  - 20

→ 10√2 u₂ = 40 × 7

→ u₂ = 280 / (10√2)

→ u₂ = 14√2   ms⁻¹

therefore,

Initial velocity of second stone will be 14 √2  ms⁻¹.