Asked by Zoomba21
An airship can travel with a horizontal speed of up to 1.30 x 10^2 km/h. If a parcel is dropped from this airship, so that it lands 135 m in front of the spot over which it was released, how far above the ground is the airship?
Answers
Answered by
drwls
First convert 130 km/h to 36.1 m/s.
The time is takes to reach the ground is
T = (horizontal distance)/(horizontal velocity)
= (135 m)/(36.1 m/s)
= 3.74 s
The altitude H of the airship can be computed from the "time to fall", T, as follows:
H = (1/2)g T^2
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