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Question: From a balloon rising vertically upward at \(5\,m/s\) a stone is thrown up at \(10\,m/s\) relative t...

From a balloon rising vertically upward at 5 m/s5\,m/s a stone is thrown up at 10 m/s10\,m/s relative to the balloon. Its velocity with respect to the ground after 2s2s is. (Take g=10 m/s2g = 10\,m/{s^2})
(A) 0 m/s0\,m/s
(B) 20 m/s20\,m/s
(C) 10 m/s10\,m/s
(D) 5 m/s5\,m/s

Explanation

Solution

Hint We will find the initial velocity of the balloon which is the addition of the initial velocity of the stone and the balloon as given in the question. Next, we will use the displacement equation S=ut+12at2S = ut + \dfrac{1}{2}a{t^2} for time t=2st = 2s to get the displacement of the stone. We will calculate the velocity using the displacement in the equation v2=u2+2aS{v^2} = {u^2} + 2aS
Formula used: Laws of motion equation
S=ut+12at2S = ut + \dfrac{1}{2}a{t^2} and
v2=u2+2aS{v^2} = {u^2} + 2aS.

Complete step by step answer
It is given that the balloon is rising vertically upwards with a velocity of 5 m/s5\,m/s. This is the initial velocity of the balloon and it does not change.
From this balloon, we throw a stone at a speed of 10 m/s10\,m/s upwards, in the same direction as that in which the balloon is travelling.
Thus the initial velocity of the stone is 15 m/s15\,m/s, since the speed of the balloon will get added to it.
Now we have to find the velocity of the balloon at the time t=2st = 2s.
For this, we will use the laws of motion equation
S=ut+12at2S = ut + \dfrac{1}{2}a{t^2},
where SS is the required displacement.
Substituting the value of t=2st = 2s and the initial velocity of u=15 m/su = 15\,m/s in the above equation, we get S=(15 m/s)(2s)+12(−10 m/s2)×(2s)2S = (15\,m/s)(2s) + \dfrac{1}{2}( - 10\,m/{s^2}) \times {(2s)^2}
⇒S=20 m\Rightarrow S = 20\,m.
Next, we will use
v2=u2+2aS{v^2} = {u^2} + 2aS
⇒v2=(15 m/s)2+2(−10 m/s2)(20 m)=(5 m/s)2\Rightarrow {v^2} = {(15\,m/s)^2} + 2( - 10\,m/{s^2})(20\,m) = {(5\,m/s)^2}
Thus the magnitude of the velocity is v=5 m/sv = 5\,m/s
This is the velocity of the stone with respect to the ground after 2s2s.

Therefore, the correct answer is option (D).

Note The initial velocity of the stone needs to be found carefully since the reference frame should be with respect to the ground. Thus we add the velocity of the balloon with respect to the ground with the velocity of the stone with respect to the balloon. Here we will add the two velocities and not subtract, since both the velocities are directed towards the same direction and not opposite or at any other angle.