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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

Verified Answer :

syed514

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.

scme1702

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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