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13 February, 12:40

A hot air balloon is traveling vertically upward at a constant speed of 2.8 m/s. When

it is 29 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

What is its speed just before impact with the

ground?

Answer in units of m/s

+2
Answers (2)
  1. 13 February, 15:07
    0
    why is your formula x=? if i'm reading it right the formula is the height of the object at time t. That being the case, set your setup (which looks right) equal to 0 and solve for t. This will tell you how long it takes to hit the ground. Plug this value of t into the forumla - 9.8t+2.8 and that will tell you the speed at time t.
  2. 13 February, 15:50
    0
    We can use the equation of motion:

    2as = v² - u²; where s = 29 m, a = 9.8 m/s², u = - 2.8 m/s

    v = √ (2 x 29 x 9.8 + 2.8²)

    v = 24. 0 m/s
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