A balloon is rising with a constant speed 10m/ s a stone is dropped from it when the balloons is at height of 40 m . Total distance covered by the stone before reaching the ground is
Answers
Answer:
Explanation:
The balloon first reaches a height of 40 m, then a stone is dropped . So the stone will also have the velocity of the balloon as they are tied together.
The stone first moves certain distance say 'x ' metres with its velocity to a point b where its velocity becomes 0 m/s . Then it again falls to the ground.
With this basic idea we will move on solving our question.
u = 10 m/s
v = 0 m/s
s = x
a =g= - 10 m/s^2 ( as it is against earth's gravity )
v^2 - u^2 = -2gx
0^2 - 10^2= - 2*10*x
-100 = - 20x
x = 100/20
x = 5m
For reaching its topmost point the stone has to travel 5 m.
Well if we apply a little common sense here we can avoid unwanted steps.
the distance it travels can never change so it would be the same.
First it travelled 'x' metres then another 'x' metres as it falls down and then the complete height of the balloon till it reaches the ground.
Total distance = x+x+ 40 m
= 2x + 40 m
= 2(5 ) + 40 m
= 10 +40 m
= 50 m