Question

A stone is allowed to fall from top of a tower of height 100m with final velocity 20m/s . At the same time another stone is projected upwards vertically with velocity of 25m/s from ground.Calculate when and where will they meet ? Plzz give step by step detail explanation​

Answer 1

Answer:

• The stones will meet after time (t) 4 seconds.

• The stones will meet at 80 m from the top of the tower.

Given:

Stone "A"

• Initial velocity (u) = 0 m/s

      ∵ [ Ball is dropped ]

• Let it travel a distance "x" before it meets.

• g = + 10 m/s² (Moving down)

Stone "B"

• Initial velocity (u) = 25 m/s.

• Let it travels 100 - x distance before it meets.

• g = - 10 m/s² (Moving up)

Explanation:

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Stone "A"

Applying Second, Kinematic Equation

⇒ S = u t + 1 / 2 a t²

Substituting the values,

⇒ x = 0 × t + 1 / 2 × 10 × t²

⇒ x = 0 + 5 × t²

⇒ x = 5 × t²

⇒ x = 5 t² __[1]

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Stone "B"

Applying Second, Kinematic Equation

⇒ S = u t + 1 / 2 a t²

Substituting the values,

⇒ 100 - x = 25 × t + 1 / 2 × (-10) × t²

⇒ 100 - x = 25 t + (-5) × t²

⇒ 100 - x = 25 t - 5 t²

⇒ 100 - x = 25 t - 5 t² __[2]

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Adding Equation [1] & [2]

⇒ 100 - x + x = 25 t - 5 t² + 5 t²

⇒ 100 = 25 t

⇒ t = 100 / 25

⇒ t = 4

⇒ t = 4 sec.

∴ The stones will meet after time (t) 4 seconds.

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Substituting the value of time (t) in Equation [1]

⇒ x = 5 t²

⇒ x = 5 × (4)²

⇒ x = 5 × 16

⇒ x = 80

⇒ x = 80 m

∴ The stones will meet at 80 m from the top of the tower.

Note:

• Symbols have their usual meanings.

• If you want distance from the ground Subtract 80 m from the total height of the tower i.e. 20 m.

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Answer 2

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the stones will meet after time 4 seconds.