A hot air balloon is travelingvertically up- ward at a constant speed of3.8m/s.W

Question

A hot air balloon is travelingvertically up-

ward at a constant speed of3.8m/s.When

it is 23 m above the ground, apackage is

released from theballoon.

The acceleration of gravity is9.8m/s2 .

After it is released, for howlong is the

package in the air? Answer inunits of s.

ground? Answer in units ofm/s.

(part 3 of 4)

Now assume the hot air balloonis traveling

vertically downward at aconstant speed of

3.8 m/s.

After the package is released,how long is it

in the air? Answer in units ofs.

(part 4 of 4)

What is its speed just beforeimpact with the

Explanation / Answer

PART 3:
xf = 0m (ground) xi = 23m vi = -3.8m/s a = -9.8m/s^2
xf = xi + vit + (1/2)a(t)^2 0 = 23 - 3.8t - 4.9 t^2 A = -4.9 B = -3.8 C = 23
Quadratic formula gives two answers:
-2.588 or 1.81 units of time must be positive, so 1.81s is your answer
xf = 0m (ground) xi = 23m vi = -3.8m/s a = -9.8m/s^2
xf = xi + vit + (1/2)a(t)^2 0 = 23 - 3.8t - 4.9 t^2 A = -4.9 B = -3.8 C = 23
Quadratic formula gives two answers:
-2.588 or 1.81 units of time must be positive, so 1.81s is your answer

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