Question

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A stone is allowed to fall from the top of a tower 100m high and at same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.

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Answer 1

Given :-

A stone is allowed to fall from the top of a tower of height = 100 m

Acceleration due to gravity = 10 m/s²

Initial velocity of stone projected upwards = 25 m/s

To Find :-

When will the two stones meet.

Where will the two stones meet.

Solution :-

We know that,

• g = Gravity
• u = Initial velocity
• t = Time
• s = Displacement

Using the formula,

$\underline{\boxed{\sf Displacement=ut+\dfrac{1}{2} gt^2}}$

Given that,

Initial velocity (u) = 0 m/s

Gravity (g) = 10 m/s

Substituting their values,

Let the height covered by the falling stone by s₁.

⇒ s₁ =  0 × t + 1/2 (10) t²

⇒ s₁  = 10/2 t²

⇒ s₁ =  5t² ____(1)

Given that,

Gravity (g) = 10 m/s

Initial velocity (u) = 25 m/s

Substituting their values,

Let the distance covered by the stone thrown upward be s₂.

⇒ s₂ = 25t + 1/2 (-10) t²

⇒ s₂ = 25t + -10/2 t²

⇒ s₂ = 25 - 5t² ____(2)

Given that,

Total height = 100 m

By adding (1) and (2),

s₁ + s₂ = 100m

⇒ 5t² + (25t - 5t²) = 100

⇒ 5t² - 5t² + 25t = 100

⇒ 25t = 100

⇒ t = 100/25

⇒ t = 4 sec ____(3)

By substituting (3) in (1),

s₁ = 5t²

⇒ s = 5 (4)²

⇒ s = 5 (16)

⇒ s = 80 m

Therefore, the two stones will meet at a distance of 80 m after 4 sec.