ICSE Class 9 Physics Question 18 of 20

Laws of Motion — Question 18

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Question

Question 18

A stone is dropped freely from the top of a tower and it reaches the ground in 4 s. Taking g = 10 m s-2, calculate the height of the tower.

Answer

From the equation of motion,

h = ut + 12\dfrac{1}{2} gt2

Given,

g = 10 m s-2

t = 4 s

u = 0

Substituting the values in the formula, we get,

h=(0×4)+(12×10×42)h=0+(5×16)h=5×16h=80 m\text h = (0 \times 4) + (\dfrac{1}{2} \times 10 \times 4^2) \\[0.5em] \text h = 0 + (5 \times 16) \\[0.5em] \text h = 5 \times 16 \\[0.5em] \text h = 80\ \text m \\[0.5em]

Hence, the height of the tower is 80 m