Asked by vonne

A hot air balloon is descending at a rate of 2.4 m/s when a passenger drops a camera.
(a) If the camera is 50 m above the ground when it is dropped, how long does it take for the camera to reach the ground? (s)
(b) What is its velocity just before it lands? Let upward be the positive direction for this problem. (m/s)
Answered by bobpursley
a) The initial velocity is -2.4m/s

hf=ho+Vi*t-4.9t^2
hf=0, hi=50, Vi given, solve for t.
b) Vf=Vi-9.8t
Answered by Heather
ok so the rate that the camera is dropped is the same as the balloon is the camera's initial velocity. the change in x is 50m and the acceleration is
-9.8m/s^2 due to gravity.

a) use the equation, change in x= Vot + 1/2at^2 to find the time for the camera to reach the ground

so you now have the time, initial velocity, and acceleration

b) use the equation V= Vo + at to find the final velocity just before it lands
Answered by ju
A hot-air balloon is descending at a rate of 1.5 when a passenger drops a camera.If the camera is 41 above the ground when it is dropped, how long does it take for the camera to reach the ground?
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