Im lost on how they are solving for T in part b A hot-air balloonist, rising ver
ID: 1274637 • Letter: I
Question
Im lost on how they are solving for T in part b
A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00m/s , releases a sandbag at an instant when the balloon is a height h = 40.0m above the ground . After it is released, the sandbag is in free fall, (a) Compute the position of the sandbag at a time 0.250s and 1.00 s after its release, (b) How many seconds after its release will the bag strike the ground? (c) With magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches? Sketch ay -1, vy -1 and y -1 graphs for the motion.Explanation / Answer
The basic equation for solving this is:
d = vt + (1/2)at2
where d is displacement traveled in a certain amount of time (t), v is starting velocity,a is acceleration (must be constant), and t is time.
starting velocity of bag will be same as that of baloonist=5
displacement=40m
acceleration=-9.8
then they calculated time with the help of it , it will be positive so t=3.41 sec is answetr