Acceleration Question?

Question

A camera falls from a balloon, hitting the ground at a speed of 223.0 m/s. While ignoring air resistance, what is the (a) height above the ground in meters from which the camera was dropped and (b) the time in seconds it took for the camera to fall.

My answers came out to:

(a) 2.538 x 10^3 m

(b) 22.76 s

Is this correct? Thanks.

Also, am I correct in using +9.8m/s^2 for this question? When would I use -9.8/ms^2 for acceleration? Thanks.

Answer 1

v = u + gt; where t is the time of fall in sec, g = 9.81 m/sec^2, u = 0 the initial velocity (assuming the balloon is not ascending or descending when you dropped your expensive Nikon), and v = 223 m/sec when your camera becomes so many electronics parts. Solve for t = v/g ~ 22 sec (using g = 10, use 9.81 when you do it).

Now h = ut + 1/2 gt^2 = 0 + 1/2 g(v/g)^2 = height the camera fell in about 22 seconds. Through cancellation etc. we have h = 1/2 v^2/g ~ 1/2 4 X 10^4/10 = 2 X 10^3 meters = balloon height when the camera slipped from your hands.

Your answers are close to my approximations.

As to g < 0 or g > 0, pick one and then be consistent. For example, if you say acceleration downward is g > 0; then acceleration upward has to be g < 0. And also, this is important, distance S and velocity u and v have to also be consistent with whatever you assign to g.

For example, in v = u + gt, I used g > 0 to point downward towards the ground. Thus, v > 0 and u > 0 mean the velocity vectors also point to the ground. As you get along in math, you will find that vectors have to be consistent, but at the outset you can pick which way is positive and by default which way is negative.

Answer 2

Since the object fell from rest you have y = gt^2/2
and v = gt. From the second eq:
223 = 9.8t or t = 22.76 secs. so y = 9.8* 22.76^2/2 =2538
So, your answers agree with mine.

Answer 3

i got a) 2537m and b)22.75 sec

+9.8 is correct