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Motion in A Straight Line question

2021 · 25 Jul · Shift 2 · Q53
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  5. /2021 · 25 Jul · Shift 2 · Q53

Motion in A Straight Line question

2021 · 25 Jul · Shift 2 · Q53

JEE MainPhysicsMotion in A Straight LineMCQ+4 / −1
A balloon was moving upwards with a uniform velocity of 10 m/s. An object of finite mass is dropped from the balloon when it was at a height of 75 m from the ground level. The height of the balloon from the ground when object strikes the ground was around : (takes the value of g as 10 m/s2)
  1. A
    300 m
  2. B
    200 m
  3. C
    125 m
  4. D
    250 m
View written solutionFree

Correct answer: C

  1. Initial conditions of the dropped object

When the object is dropped from the balloon:

  • Initial height from ground: y0=75 my_0 = 75\,\text{m}y0​=75m
  • Initial upward velocity: u=10 m/su = 10\,\text{m/s}u=10m/s
  • Acceleration due to gravity: a=−10 m/s2a = -10\,\text{m/s}^2a=−10m/s2

Taking upward as positive, the position of the object after time ttt is

y=y0+ut+12at2y = y_0 + ut + \frac{1}{2}at^2y=y0​+ut+21​at2

So,

y=75+10t−5t2y = 75 + 10t - 5t^2y=75+10t−5t2

The object strikes the ground when y=0y = 0y=0.

Thus,

75+10t−5t2=075 + 10t - 5t^2 = 075+10t−5t2=0

Divide by −5-5−5:

t2−2t−15=0t^2 - 2t - 15 = 0t2−2t−15=0

Factorizing:

(t−5)(t+3)=0(t-5)(t+3)=0(t−5)(t+3)=0

Hence,

t=5 st=5\,\text{s}t=5s

(Reject t=−3 st=-3\,\text{s}t=−3s as unphysical.)

  1. Position of the balloon after 5 s

The balloon keeps moving upward with uniform velocity 10 m/s10\,\text{m/s}10m/s.

In 5 s5\,\text{s}5s, it rises by

Δh=vt=10×5=50 m\Delta h = vt = 10 \times 5 = 50\,\text{m}Δh=vt=10×5=50m

So its height from the ground becomes

h=75+50=125 mh = 75 + 50 = 125\,\text{m}h=75+50=125m
  1. Checking options
  • A: 300 m300\,\text{m}300m ❌
  • B: 200 m200\,\text{m}200m ❌
  • C: 125 m125\,\text{m}125m ✅
  • D: 250 m250\,\text{m}250m ❌

Therefore, the correct answer is Option C.

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