A mass falls a height H such that it covers half the height in the last second. What is the the height H and the total time taken for the fall?
The acceleration due to gravity of a body which is falling freely is not affected by its mass. The acceleration is usually taken to be a constant equal to 9.8 m/s^2.
Let the height that the body falls be equal to H. Now we are given that it covers half of this height in the last second.
When the body falls its velocity is 0 m/s. This increases gradually as the body falls. The relation between the distance travelled, the acceleration and the initial velocity is given by:
d = (2u + at)*(t/2), where d is the distance travelled, u is the initial velocity when the free fall started, a is the acceleration and t is the time being considered.
Let the total time the body falls be T seconds.
The first H/2 takes T - 1 s and the last H/2 takes 1 s
So we get H/2 = (0 + 9.8*(T-1))(T-1)/2
and H/2 = (2*9.8*(T - 1) + 9.8*1)(1/2)
H/2 = (0 + 9.8*(T-1))(T-1)/2
=> H = 9.8*( T - 1)^2
H/2 = (2*9.8*(T - 1) + 9.8*1)(1/2)
=> H = 19.6T - 19.6 + 9.8
=> H = 19.6T - 9.8
9.8*(T - 1)^2 = 19.6T - 9.8
=> (T - 1)^2 = 2T - 1
=> T^2 + 1 - 2T = 2T - 1
=> T^2 - 4T + 2 = 0
Solving the quadratic equation we get T = 3.414 s and .5857 s, we can ignore the second root.
Therefore the time for the fall is 3.414 s.
Now H = 19.6T - 9.8
=> H = 19.6*3.414 - 9.8 = 57.11m
Therefore we get the distance it falls as 57.11 m and the time taken as 3.414 s.
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