An elevator (in which a man is standing) is moving upwards with a constant acceleration of 2 m/s2. The man drops a coin from a height of 1.5 m. Find the time taken by the coin to reach the floor :- (g = 10 m/s2)1 sec
Question
An elevator (in which a man is standing) is moving upwards with a constant acceleration of 2 m/s2. The man drops a coin from a height of 1.5 m. Find the time taken by the coin to reach the floor :- (g = 10 m/s2)1 sec
Solution
To solve this problem, we need to use the second equation of motion which is:
s = ut + 0.5at²
where: s = distance (1.5 m) u = initial velocity (0 m/s, because the coin is dropped) a = acceleration (the sum of the acceleration of the elevator and gravity, which is 2 m/s² + 10 m/s² = 12 m/s²) t = time (which we are trying to find)
Substituting the known values into the equation, we get:
1.5 = 0t + 0.512*t²
Solving for t² gives us:
t² = 1.5 / (0.5*12) t² = 1.5 / 6 t² = 0.25
Taking the square root of both sides to solve for t gives us:
t = sqrt(0.25) t = 0.5 seconds
So, the time taken by the coin to reach the floor is 0.5 seconds.
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