Question 4: Motion in 1D

Question

A gas balloon is going up with a constant velocity of $10 \mathrm{~m} / \mathrm{s}$. 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 $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

When an object is dropped from a moving carrier (like the balloon), it gets the velocity of that carrier at the exact moment of release.
Initial velocity of the stone (balloon) is $\mathrm{u}_{\mathrm{s}}=10 \mathrm{~m} / \mathrm{s}$ (upwards).
Initial height of the stone is $\mathrm{h}=75 \mathrm{~m}$.
The acceleration due to gravity is $\mathrm{g}=-10 \mathrm{~m} / \mathrm{s}^2$ (downwards).
Using the second equation of motion for the stone's displacement along vertical direction,
$$ \mathrm{y}=\mathrm{u}_{\mathrm{y}} \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{y}} \mathrm{t}^2 $$
Since the stone eventually hits the ground (at height 0 ), its displacement from the starting point ( 75 m ) is $\mathrm{y}=$ -75 m .
$$ -75=(10) \mathrm{t}+\frac{1}{2}(-10) \mathrm{t}^2 $$
$$\Rightarrow $$ $-75=10 t-5 t^2$
$$\Rightarrow $$ $15=-2 t+t^2$
$$\Rightarrow $$ $t^2-2 t-15=0$
$$\Rightarrow $$ $$ (t-5)(t+3)=0 $$
Since time cannot be negative, we take $\mathrm{t}=5$ seconds.
While the stone is falling for 5 seconds, the balloon continues to move upward at a constant velocity.
The constant velocity of balloon is $\mathrm{v}_{\mathrm{b}}=10 \mathrm{~m} / \mathrm{s}$.
The time interval of journey is $\mathrm{t}=5$ seconds.
Displacement $=$ Velocity ÃÂ Time
Displacement $=10 	imes 5=50 \mathrm{~m}$
The final height is the initial height plus the displacement covered during the 5 seconds.
Total Height $=75 \mathrm{~m}+50 \mathrm{~m}=125 \mathrm{~m}$
Therefore, the height of the balloon when the stone hits the ground is 125 m .
Hence, the correct option is (D).

Answer

85

Answer

150

Answer

129

Answer

125

Motion in 1D

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