A hot-air balloonist, rising vertically with a constant speed of 5.00 m/s releases a sandbag at the instant the balloon is 40.0 m above the ground. After it is released, the sandbag encounters no appreciable air drag. Compute the position of the sandbag at 0.250 s after its release.

40.9 m

39.7 m

41.3 m

40.3 m

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luisejr77 luisejr77 · Answer 1 · 2019-04-15T21:38:04+02:00

Answer:

First option

The position of the sandbag at 0.250 s after its release is 40.9 m

Explanation:

We want to know the height of the balloon as a function of time, after 0.250s

Then we use the following equation of motion:

[tex]h(t) = h_0 + s_0t - 0.5gt ^ 2[/tex]

In this equation

[tex]s[/tex] is the speed

[tex]g[/tex] is gravitational acceleration

[tex]h_0[/tex] is the initial height

[tex]s_0[/tex] is the initial velocity.

If the sandbag is released 40 meters above the ground, then the initial height is 40 meters.

As the balloon rises at a constant speed, then the initial velocity of the balloon is 5 m/s

Now we substitute these values in the formula

[tex]h(t) = 40 + 5t - 0.5(9.8)t ^ 2\\\\h(t) = 40 + 5t - 4.9t ^ 2[/tex]

Now we look for:

[tex]h(t=0.250\ s)[/tex]

[tex]h(0.250) = 40 +5(0.250) - 4.9(0.250) ^ 2\\\\h(0.250) = 40.9\ m[/tex]