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Question

 A balloon starts rising from the ground with an acceleration of $$1.25 \ m/s^2$$. After 8 seconds, a stone is released from the balloon. After releasing, the stone will


A
cover a distance of 40 m till it strikes the ground.
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B
have a displacement of 50 m till it reaches the ground
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C
reach the ground in 4 seconds.
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D
begin to move down instantaneously
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Solution

The correct option is C reach the ground in 4 seconds.
The velocity of balloon at time $$t=8$$sec
$$v=u+at$$
Initial velocity$$=u=0$$
Acceleration$$=a=1.25㎧$$
$$\therefore v=at$$
$$=1.25\times 8$$
$$v=10㎧$$
To find distance
$$2as={ v }^{ 2 }-{ u }^{ 2 }$$
$$\therefore S=\cfrac { { v }^{ 2 }-{ u }^{ 2 } }{ 2a } $$
$$=\cfrac { 100 }{ 2\times 1.25 } $$
$$\therefore S=40m$$
After stone release it will cover same distance in opposite direction
$$\therefore S=-40m$$
Free fall of motion
$$\therefore S=ut-\cfrac { 1 }{ 2 } g{ t }^{ 2 }$$
$$-40=10t-5{ t }^{ 2 }$$
On solving we get two values of t
$$t=-2$$ and $$t=4$$
Time cannot be negative
$$\therefore $$ stone will reach the gound in $$4$$ sec

Physics

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