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

From a balloon rising vertically upwards at 5m/s5m/s a stone is thrown up at 10m/s10m/s relative to the balloon. Its velocity with respect to ground after 2 s is (take g=10m/s2g=10m/{{s}^{2}} )
A. 0m/s B. 20m/s C. 10m/s D. 5m/s \begin{aligned} & \text{A}\text{. }0m/s \\\ & \text{B}\text{. }20m/s \\\ & \text{C}\text{. }10m/s \\\ & \text{D}\text{. }5m/s \\\ \end{aligned}

Explanation

Solution

To solve this question, first find the relative velocity of the stone with respect to the ground. Obtain the laws of motion to find the final velocity. Then using the laws of motion find the final velocity of the stone by using the given values.

Complete step by step answer:
The velocity of the balloon rising vertically upward with respect to the ground is, vb=5m/s{{v}_{b}}=5m/s
The velocity of the stone thrown upward relative to the balloon is, vs=10m/s{{v}_{s}}=10m/s
So, the velocity of the stone with respect to the ground will be,

u=vb+vs u=5m/s+10m/s u=15m/s \begin{aligned} & u={{v}_{b}}+{{v}_{s}} \\\ & u=5m/s+10m/s \\\ & u=15m/s \\\ \end{aligned}

So, he initial velocity of the stone with respect to the ground will be u=15m/su=15m/s
Let, after 2 seconds, the final velocity of the stone will be v.
So, we can write that,
v=u+atv=u+at
Now, the acceleration acting on the stone is the acceleration due to gravity and the direction of the acceleration is opposite to the direction of the velocity of the stone. So, we can write acceleration as,
a=g=10m/s2a=-g=-10m/{{s}^{2}}
Putting the values on the above equation, we get that,

v=1510×2 v=5m/s \begin{aligned} & v=15-10\times 2 \\\ & v=-5m/s \\\ \end{aligned}

So, the final velocity of the stone after 2 seconds will be 5m/s5m/s .
Hence, the correct answer is option D.

Note:
Concept of relative velocity gives different values of velocity for the same object for different observers in different frames of reference. For example, when we throw a ball inside a moving bus, the ball will move normally for the observer inside the bus. But for an observer outside the bus at rest, will see the ball moving at a speed higher than the speed of the bus.