A balloon is rising vertically upwards with a velocity of 29 ms-1. A stone is dropped from it, which reaches the ground in 10 seconds. The height of the balloon when the stone was dropped from it is (g = 9.8 ms-1)
Answers
Let the height from which the stone is released be h.
The upward velocity u of the stone when dropped from height h = 10 m/s.
The further height d to which the stone will rise further can be found from the relation;
v² - u² = 2 g d
0² - 10² = - 2× 10×d; ==> -20 d = -100, or d = 5 m.
Time taken to go up 5m can be found using
v = u + a t. At the highest point v=0 m/s, u = 10 m/s, t =1 s.
The stone starts to fall from a height (h +5)m, with an initial velocity zero, and an acceleration +10 m/s², and in time t =11 s -1s = 10s reaches the ground. Using the relation
s = u t + ½ a t², we can find s.Substituting for u, a and we get s, we get
s = 0×10 + ½ ×10×10² = 500m
s = h + 5 = 500 m; ==> h = 495 m.
The height of the balloon when the stone was dropped from it = 495 m.
Method 2:
Let the point from which the stone is dropped be taken as origin of coordinate system, upwards direction is taken as positive and downward direction is taken as negative.
Let the stone be released from the balloon at a height h above the geound, when the stone reaches the ground its displacement is -h. an upward velocity u = +10 m/s², and an acceleration due to gravity = -10 m/s², t = 11 second
-h = 10×11 -½ × 10× 11²= 110 - 5×121= 110 - 605 = - 495 m.
ie h= 495 m.
ie the height from where the stone is dropped = 495 m.