# What acceleration to use in this problem?

A balloon starts rising from the ground with an acceleration of 1.25 m/second square. After 8 second, a stone is released from the balloon. Find the time for stone to reach ground, distance covered, height of balloon when stone hit ground.

Should I take acceleration as 10 $m/s^2$ or 1.25 $m/s^2$ to find the stone's height?

There are two parts to this problem.

In the first part, the stone is rising with the balloon. It has a certain acceleration for a certain time. At the end of that it will have reached a certain height and velocity. Calculate it.

Then the stone is released. With the initial velocity and height calculated above, it now starts dropping. It is now subject to the acceleration of gravity.

With the above guidance you should be able to solve the problem. You just need to realize that there isn't a "single" acceleration throughout. So leave the math behind for a minute, and draw yourself a picture of what is going on...

I drew the picture above but deliberately didn't annotate it. This is the height of the stone with time - note it starts by accelerating up (1.25 m/s^2), then falls at t=8 (with a = -9.891 m/s^2). See how the curvature changes?

• What about the acceleration of the stone at that height? – user116688 Jul 17 '17 at 17:23
• At height=40 m I mean – user116688 Jul 17 '17 at 17:23
• Not sure I understand your question. The acceleration changes at that moment in time (I assume you are taking g=10 m/s/s ) but since the change is abrupt it doesn't matter what value you use - just that it is one value before and another value after t=8 s – Floris Jul 17 '17 at 18:45

Someone please correct me if I'm wrong, I'm new..

The balloon is rising from the groung, so if you think about it, you have to calculate the speed the balloon reaches after 8 seconds and take it as the initial speed of the stone, then since we assume for simplicity that the air resistance is not significant, there is no other force acting on the stone, just it's weight, hence, since it's weight is the only force acting on the stone, take $10m/s^2$

a = 1.25m/s^2

Now , v = u + at

= 1.25 x 8 = 10 m/s

Hence s = 1/2 at^2 = (1.25)(64)/2

= 40 m

Now, H= -40m ( due to downward direction.)

and 40 = 10t - 5t^2 ( s= ut +1/2 at^2)(placing above obtained and given values)

therefore , t^2 - 2t + 8 = 0

**t = 4**
• Welcome to Physics! Note that this site has MathJax enabled, which means you can use Latex-like syntax to add in equations for readability. Note that answers this terse are not very useful as it provides little insight to the problem. – Kyle Kanos Aug 8 '15 at 18:15