A hot air balloon is traveling vertically upward at a constant speed of 3.7 m/s.

Question

A hot air balloon is traveling vertically upward at a constant speed of 3.7 m/s. When
it is 17 m above the ground, a package is
released from the balloon.
After it is released, for how long is the
package in the air? The acceleration of gravity
is 9.8 m/s
2
.
Answer in units of s

part 2 : What is its speed just before impact with the
ground?
Answer in units of m/s

part 3: Now assume the hot air balloon is traveling
vertically downward at a constant speed of
3.7 m/s.
After the package is released, how long is it
in the air?
Answer in units of s

Part 4: What is its speed just before impact with the
ground?
Answer in units of m/s

Explanation / Answer

(2 x 3.7)/9.8 = 0.755 secs 3.7^2 + (2 x 9.8 x 17) = v^2 = 346.89 sq-root = v = 18.62 m/s 17/((3.7 + 18.62)/2) = 1.523 secs 0.755 + 1.523= 2.278 secs

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