A balloonist, riding in the basket of a hot air balloon that is rising vertically with a constant velocity of 10.1 m/s, releases a sandbag when the balloon is 43.4 m above the ground. Neglecting air resistance, what is the bag's speed when it hits the ground? Assume g = 9.80 m/s2.
m/s
5 answers
Larry, or Jake: I will be happy to critique your thinking.
Tried using a free fall set up with no luck. Things like (-9.8)(10.1)+43.3
You can get a quick answer using energy conservation.
(1/2)M V2^2 = M g H + (1/2) M V1^1
M's cancel.
V2 = sandbag when it hits the ground
V1 = sandbag velocity when released (relative to ground) = 10.1 m/s
H = 43.4 m
(1/2)M V2^2 = M g H + (1/2) M V1^1
M's cancel.
V2 = sandbag when it hits the ground
V1 = sandbag velocity when released (relative to ground) = 10.1 m/s
H = 43.4 m
30.87 m/s
answwr
1 answer