A hot-air balloon is rising upward with a constant speed of 2.45 m/s. When the balloon is 3.22 m above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?
I know that average speed equals distance / elapsed time and the question gives the speed and distance so would I set this up as 2.45m/s^ = 3.22m / T ?
You are absolutely on the wrong track.
Compass:
vi=2.45m/s
hi=3.22
hf=0
hf=Hi+vi*t-4.9t^2 solve for t. Notice it is a quadratic.
When compass begins to move its velocity is directed upwards and it moves until the velocity becomes zero.
hₒ =vₒ•t1-g•t1²/2
v=vₒ -gt1.
0 = vₒ -g•t1
t1= vₒ/g =2.45/9.8 =0.25 s.
hₒ =vₒ²/2•g =2.45²/2•9.8 =0.306 m
H =hₒ + h = 0.306+3.22 =3.526 m.
Downward motion
H =g•t2²/2
t2=sqrt(2•H/g) = sqrt (2•3.526/9.8) =0.85s.
t =0.25 +0.85 = 1.1 s.
To solve the problem, we need to find the elapsed time (T) it takes for the compass to hit the ground. Here's how we can set up the equation:
Average speed = distance / elapsed time.
In this case, the average speed is given as 2.45 m/s, and the distance the compass has to travel is 3.22 m. So, the equation becomes:
2.45 m/s = 3.22 m / T.
To find the elapsed time (T), we can rearrange the equation:
T = 3.22 m / 2.45 m/s.
Simplifying the right side, we get:
T ≈ 1.3163 s.
Therefore, it takes approximately 1.3163 seconds for the compass to hit the ground after it is dropped from the hot-air balloon.