Question

A stone fall from a building and reaches the ground 2.5 second later. How high is the building? (g=9.8 m/s^2.

Answer 1

Answer :-

Given,

• Initial Velocity (u) = 0m/s
• Acceleration due to gravity (g) = 9.8 m/s
• Time taken = (t) = 2.5 seconds

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$\bf {h = ut + \frac{1}{2} {gt}^{2}}$

➟ h = (0 × 2.5 s) + 1/2 × 9.8 m/s × 2.5s × 2.5s

➟ h = 0 + 4.9 × 2.5s × 2.5s

➟ h = 4.9 × 6.25

➟ h = 30.625m

The height of the building is 30.625 m.

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Answer 2

✬ Answer:- ✬

Given:-

• Time taken (t) = 2.5 s
• Initial velocity (u) = 0 m s⁻¹
• Acceleration due to gravity (g) = 9.8 m s⁻²

To Find:-

Height of the Building.

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We know,

➯ h = ut + ½gt²

where,

1. h = Distance, (height)
2. u = Initial velocity,
3. t = Time taken
4. g = Acceleration due to gravity.

Putting the values,

h = (0 m/s)t + ½(9.8 m/s²)(2.5 s)²

⟹ h = ½ • 9.8 m/s² × 6.25 s²

⟹ h = 4.9 m × 6.25

⟹ h = 30.625 m

(⟹ h ≈ 31 m)

∴ Height of the building was 30.625 metres.