Problem 2: A helicopter is ascending vertically at a speed of 19.6 m/s. 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?
Strategy
Initially, the stone is part of the helicopter and is 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. Then it starts falling 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.
Solution
- 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 s … (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
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 s … (2)
- Total time
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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What if we solve this problem simply by putting values in the equation “time= distance / speed”?
Sana
Formula t = s/v is applicable only in the case when a body is moving with uniform velocity. Here the stone is continuously under the influence of gravity and the velocity is changing every moment.
Nice
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