When does the stone hit the ground after being dropped from balloon at 75m moving up at 10m/s?

t=5s. From y=75+10t−5t²=0: t²−2t−15=0, giving t=5s.

What is the balloon height when stone hits ground?

125m. Balloon at constant 10m/s: 75+10×5=125m.

Gas balloon at 75m going up at 10 m/s, stone dropped — height of balloon when stone hits ground

JEERANKERS

PhysicsKinematics

QA gas balloon is going up with a constant velocity of 10 m/s. When this balloon reached a height of 75 m, a stone is dropped from it and balloon keeps moving up with the same velocity. The height of the balloon when the stone hits the ground is ________ m. (Take $g = 10$ m/s²)

A) 85 B) 150 C) 129 D) 125Kinematics

MCQ

✅ Correct Answer

D) 125 m

✓

Solution

Set up equations for stone

At $t=0$: stone is at height 75 m with initial velocity $+10$ m/s upward (same as balloon).

$y_{\text{stone}}=75+10t-\dfrac{1}{2}(10)t^2=75+10t-5t^2$

Find time for stone to hit ground

$y_{\text{stone}}=0$:

$75+10t-5t^2=0$
$5t^2-10t-75=0$
$t^2-2t-15=0$
$(t-5)(t+3)=0$

$t=5\text{ s}$ (taking positive value)

Find balloon height at t = 5s

$y_{\text{balloon}}=75+10\times5=75+50=\boxed{125\text{ m}}$

📘 Kinematics of Dropped Object from Moving Body

When an object is dropped from a moving body, it retains the velocity of the body at the moment of release. Here the stone has initial velocity +10 m/s upward (not zero) when released. The balloon continues at constant 10 m/s while the stone follows projectile motion.

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Important Concepts

Initial Velocity at ReleaseThe stone has the SAME velocity as the balloon at the moment of release: +10 m/s upward. This is a common mistake — students often assume the stone is dropped from rest.

Projectile Equationy = y₀ + v₀t − ½gt². With y₀=75m, v₀=+10m/s, g=10m/s²: y=75+10t−5t²=0 gives t=5s (taking positive root).

Balloon HeightBalloon moves at constant 10 m/s: height at t=5s = 75+10×5=125m. Note the balloon doesn't decelerate — it maintains constant velocity.

Time AnalysisThe stone first rises for 1s (reaches 75+10=80m), then falls for 4s to ground. Total time = 5s. Balloon: 75+50=125m.

FAQs

Why does the stone have initial velocity 10m/s upward?

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When dropped from the balloon, the stone retains the balloon's velocity at that instant. The balloon (and stone before release) was moving at 10 m/s upward.

How to solve t²−2t−15=0?

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Factor: (t−5)(t+3)=0. Roots: t=5 and t=−3. Since t must be positive, t=5s.

How high does the stone go before falling?

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Stone rises until v=0: 10−10t=0, t=1s. Height at t=1s: 75+10−5=80m. Then falls from 80m to ground.

What if initial velocity were zero?

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Then y=75−5t²=0, t=√15≈3.87s. Balloon height would be 75+38.7=113.7m. The initial upward velocity significantly changes the answer.

Is this from JEE Main 2026?

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Yes, this appeared in JEE Main 2026.

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