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Problem 3: A helicopter is ascending vertically at a speed of 19.6 ms-1. When it is at a height of 156.8 m above the ground, a stone is dropped. How long does the stone take to reach the ground?

Solution

Strategy: Initially, the stone is part of the helicopter and moving up with the same speed. Hence, when it is released it goes up with the same speed until its velocity becomes zero due to the downward gravitational force. When its velocity becomes zero, it would have covered some distance to start fall down. Therefore, we have to calculate two time intervals; one, when it goes up to reach the maximum height, and two, to reach the ground from that highest point.

1. For upward motion,

Initial velocity = vi = 19.6 ms-1, Final velocity = vf = 0 ms-1, Acceleration = g= – 9.8 ms-2,  t = ?

Apply equation of motion, vf = vi + gt, we have

0 = 19.6 – 9.8t  Or 9.8t = 19.6 Or t = 19.6 ÷ 9.8 = 2 sec   …    (1)

1. Downward motion

So, the stone goes up for 2 sec. The distance covered in these 2 sec can be found by

2gs = vf2 – vi2. Put the values,

2(-9.8)s = 0 – (19.6)2 Or s = (19.6 * 19.6)/(2 * 9.8) Or s = 384.16/19.6 = 19.6 m

1. Total time

So, the total distance the stone drops through is (156.8 + 19.6) = 176.4 m under the action of gravity. Therefore, to find time t, we apply, s = vit + ½ at2, where vi = 0. Put the values,

176.4 = 0 + ½ * 9.8 * t2 Or 176.4 = 4.9t2 Or t2 = (176.4)/4.9 = t2 = 36 Or t = 6 … (2)

Total time taken to the ground is the sum of times in equation (1) and (2). Therefore, the stone will reach the ground in 2 + 6 = 8 seconds.