**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.

**For upward motion,**

Initial velocity =
v_{i} = 19.6 ms^{-1}, Final velocity = v_{f} = 0 ms^{-1},
Acceleration = g= – 9.8 ms^{-2},
t = ?

Apply equation of
motion, v_{f} = v_{i} + gt, we have

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

**Downward motion**

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

2gs = v_{f}^{2}
– v_{i}^{2}. 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

**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 = v_{i}t + ½ at^{2},
where v_{i} = 0. Put the values,

176.4 = 0 + ½ * 9.8
* t^{2} Or 176.4 = 4.9t^{2} Or t^{2} = (176.4)/4.9 = t^{2}
= 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.

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