JEE Main 2026 — Motion In One Dimension Question with Solution

When the stone is dropped, it acquires the initial velocity of the balloon. Taking the upward direction as positive, the initial velocity of the stone is $u=10$ m/s.

The displacement of the stone when it hits the ground is $S=-75$ m, and its acceleration is $a=-g=-10$ m/s${}^2$.

Using the equation of motion $S=ut+\frac{1}{2}a{t}^2$:

$-75=10t-\frac{1}{2}(10)t^2$

$-75=10t-5t^2$

$5t^2-10t-75=0$

$t^2-2t-15=0$

$(t-5)(t+3)=0$

Since time cannot be negative, $t=5$ s.

During this time, the balloon continues to move upwards with a constant velocity of $10$ m/s. The distance travelled by the balloon in $5$ s is:

$d=v\times t=10\times 5=50$ m

The total height of the balloon when the stone hits the ground is:

$H=75+50=125$ m

Answer: $125$