A stone dropped from the top of a building, takes 4 seconds to reach the ground. Find the height of the building.

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Solution

Given,
Initial velocity = u = 0
time taken = t = 4s
acceleration = g = 9.8 m/s2
Using 2nd equation of motion we get,
s=u $\times$ t + $\frac{1}{2} \times a \times t^{2}$
=(0)(4)+(1/2)(9.8)(4)(4)
=78.4 m
So, the height of the building is 78.4 m.

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A stone dropped from the top of a building, takes 4 seconds to reach the ground. Find the height of the building.

Given,Initial velocity = u = 0time taken = t = 4sacceleration = g = 9.8 m/s2Using 2nd equation of motion we get,s=u×t + 12×a×t2=(0)(4)+(1/2)(9.8)(4)(4)=78.4 mSo, the height of the building is 78.4 m.

Given,
Initial velocity = u = 0
time taken = t = 4s
acceleration = g = 9.8 m/s2
Using 2nd equation of motion we get,
s=u $\times$ t + $\frac{1}{2} \times a \times t^{2}$
=(0)(4)+(1/2)(9.8)(4)(4)
=78.4 m
So, the height of the building is 78.4 m.