A classical mechanics problem by Ankita Mehra
Classical Mechanics
Level
2
A balloon is ascending vertically with an acceleration of 0.2 m/s^2. Two stones are dropped from it at an interval of 2 sec. Find the distance between them 1.5 seconds after the second stone is released. (Use g= 9.8 m/s^2)
50
49
40
35
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1 solution
A nice shortcut is to view the situation from the rest frame of the balloon. From its perspective, the gravitational acceleration appears to be g ′ = g + a = 9 . 8 + 0 . 2 = 1 0 m/s 2 .
We must compare the distance traveled by the stones at 3.5 s (first) and 1.5 s (second). Their positions this time, relative to the balloon, are y 1 = 2 1 ⋅ 1 0 ⋅ ( 3 . 5 ) 2 = 6 1 . 2 5 m , y 2 = 2 1 ⋅ 1 0 ⋅ ( 1 . 5 ) 2 = 1 1 . 2 5 m . The difference between these heights is 5 0 m.