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Throwing and Catching at Same Height

From a balloo...

Question

From a balloon rising vertically upwards $5 m s^{- 1}$ at stone is thrown up at $10 m s^{- 1}$ relative to the balloon. Its velocity wit respect to ground after $2 s$ is (assume, $g = 10 m s^{- 2}$ ).

zero

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$- 5 m s^{- 1}$

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$10 m s^{- 1}$

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$20 m s^{- 1}$

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Solution

The correct option is B
$- 5 m s^{- 1}$

Step 1. Given data
The velocity of the balloon $= 5 m s^{- 1}$
The velocity of the stone $= 10 m s^{- 1}$
Time, $t = 2 s$
Initial velocity, $u$
$u =$ velocity of balloon $+$ velocity of stone
$u = 10 + 5$
$u = 15 m s^{- 1}$
Step 2. Finding the velocity of stone with respect to the ground, $v$
By using the Equation of motion,
$v = u + g t$ [Where, $u$ is the initial velocity]
$v = 15 + (- 10 \times 2)$ [Where, $g = - 10 m s^{- 2}$ as the stone is thrown upward.]
$v = - 5 m s^{- 1}$
Hence, the correct option is $(B)$ .

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From a balloon rising vertically upwards 5ms-1 at stone is thrown up at 10ms-1 relative to the balloon. Its velocity wit respect to ground after 2s is (assume,g=10ms-2).

From a balloon rising vertically upwards $5 m s^{- 1}$ at stone is thrown up at $10 m s^{- 1}$ relative to the balloon. Its velocity wit respect to ground after $2 s$ is (assume, $g = 10 m s^{- 2}$ ).