A small object is thrown out of a hot air balloon from an unknown height. It falls straight down and hits the ground 5.6 seconds later with a speed of 125 m/s. How high is the balloon?

Question

shane_b · Answer

Information given:$$V_f=125m/s$$$$a=9.8m/s^2$$$$t=5.6s$$Since you only have constant acceleration (due to gravity) you can use the kinematic equations (see image).

First, find the initial velocity at which the object was thrown downward:
$$\large V_f=V_i+at$$$$\large V_i=V_f-at=125m/s-(9.8m/s^2)(5.6s)=70.12m/s$$
Now calculate the distance:$$\large V_f^2=V_i^2+2ad$$$$\large d=\frac{V_f^2-V_I^2}{2a}=546.336m$$

anonymous · Answer

I looked at this as a conservation of energy problem. That is:$$mgh=\frac{1}{2}mv^2$$The masses cancel out, leaving you with $$9.8\frac{m}{s^2}*h=\frac{1}{2}*(125\frac{m}{s})^2$$But I got:$$h=797m$$

shane_b · Answer

The problem is that the ball not only had PE at the top...it also had KE in the beginning since it was thrown downward.

anonymous · Answer

Got bumped. Crummy time for an update. :|
Anyhow, yes, Shane_B, that makes sense. Doing this after your first step gives me the same answer:$$mgh + \frac{1}{2}m(70.12\frac{m}{s})^2=\frac{1}{2}m(125\frac{m}{s})^2$$

shane_b · Answer

Yea...I thought I broke OpenStudy again :)

anonymous · Answer

:)

anonymous · Answer

thank you guys so much :D