A hot air balloon is rising with an acceleration of 10 $m / s^{2}$ . The moment when its velocity is 10 m/s, a stone is dropped which of the following is the displacement time graph of the stone with respect to the balloon.

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Solution

The correct option is B

Okay so what we know when stone is dropped :
$u_{b a l l o o n} = + 10 m / s$
$a_{b a l l o o n} = + 10 m / s^{2}$
$u_{s t o n e} = + 10 m / s$ by now, think you know why velocity of stone is velocity of balloon
$a_{s t o n e} = - 10 m / s^{2}$
To find $s_{s / b}$
$s = u t + \frac{1}{2} a t^{2}$
$u_{s / b} = 10 - 10 = 0$
$a_{s / b} = - 10 - 10 = - 20 m / s^{2}$
$s_{s / b} = - 10 t^{2}$

Now, Draw s on y - axis
f on x-axis
How will it look?
y = -10 $x^{2}$
Well if you don't then no problem. Be smart. Substitute some valves and draw the graph and see.

So the graph should look like this.

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