Stone falling from a Balloon - Physics

Classical Mechanics Level 2

A balloon is moving vertically upward with a velocity of 4 m/s. When it is at a height of h , a stone is dropped from it. If it reaches the ground in 4 seconds , the height of the balloon, when the stone was released , is ? Take (g = 9.8 m/s2)

62.4 m 78.4 m 42.4 m 82.2 m

Shivansh Tripathi by Shivansh Tripathi , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Karthik Rocks
Jun 27, 2014

simply use the equation s=ut+1/2gt^2 where taking all the downward forces to be positive while upward be taken as negative 

   u= - 4m/s (as ballon is moving upwards)
   t= 4 sec
   g = 9.8m/s^2

s= -4 * 4 + 1/2 * 9.8 *16

-16 + 78.4

s = 62 . 4

1 Helpful 0 Interesting 0 Brilliant 0 Confused

why u is considered as negative when moving up?? pls explain

Vandetta Jack - 6 years, 2 months ago
Jaivir Singh
Jun 27, 2014

INITIAL VELOCITY IS 4 m/s time of flight is 4 s

0 Helpful 0 Interesting 0 Brilliant 0 Confused

But this is the displacement of balloon during the time interval in which the stone touched the ground. The height of the balloon can be found by the formulae 2 h g = t \sqrt{ \frac{2h}{g}} = t Substituting t = 4; g = 9.8 we get h = 78.4. Why do you think otherwise has the value of g provided in the question?

Umang Vasani - 6 years, 11 months ago

S=ut-1\2gt^2 gives u solution:)

kajubojula Sreekanth - 6 years, 11 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...