Drops of water fall from the roof of a building 16 m high at regular intervals of time. When the first drop reaches the ground, at the same instant fifth drop begins to fall. What are the distances of the second and third drops from the roof ?6 m and 2 m 9 m and 4 m8 m and 2 m 4 m and 2 m
Question
Drops of water fall from the roof of a building 16 m high at regular intervals of time. When the first drop reaches the ground, at the same instant fifth drop begins to fall. What are the distances of the second and third drops from the roof ?6 m and 2 m 9 m and 4 m8 m and 2 m 4 m and 2 m
Solution
The problem can be solved using the equation of motion: h = ut + 0.5gt^2, where h is the height, u is the initial velocity, t is the time, and g is the acceleration due to gravity.
Given that the height is 16m and the acceleration due to gravity is approximately 9.8m/s^2, we can calculate the time it takes for the first drop to reach the ground. Since the drop is falling from rest, the initial velocity (u) is 0.
So, 16 = 0t + 0.59.8*t^2. Solving for t, we get t = sqrt(16 / 4.9) = 1.8s.
This means that the drops are falling at intervals of 1.8/4 = 0.45s.
To find the distances of the second and third drops from the roof when the first drop hits the ground, we can use the same equation of motion, but this time we know the time and we're solving for the height.
For the second drop, the time is 0.45s less than for the first drop, so t = 1.8 - 0.45 = 1.35s. Plugging this into the equation gives h = 0.59.81.35^2 = 8.8m.
For the third drop, the time is 0.9s less than for the first drop, so t = 1.8 - 0.9 = 0.9s. Plugging this into the equation gives h = 0.59.80.9^2 = 4m.
So, the second and third drops are approximately 9m and 4m from the roof, respectively.
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