A block of mass 1 kg is dropped from the top of a building of height X m. What will be the value of X if the block hits the ground with a velocity of 20 m/s? (Take the value of g = 10 m/s2)
20 m
For a freely falling body dropped from rest, use the equation of motion \(v^2 = u^2 + 2gX\), where u is the initial velocity, v the final velocity, g the acceleration due to gravity and X the height.
Since the block is dropped, the initial velocity u = 0. Given v = 20 m/s and g = 10 m/s2.
Substituting: \((20)^2 = 0 + 2 \times 10 \times X\), so \(400 = 20X\).
Solving gives \(X = \dfrac{400}{20} = 20\) m. The mass of 1 kg is not needed, since the height for a given speed does not depend on mass.
Hence, the height X of the building is 20 m.
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