A hot air balloon hovers at 92.4 m above the ground. The pilot drops a box over the edge of the basket. How fast does the person standing 10 m away, have to go to catch the box, before it hits the ground?

In order to catch the box before it hits the ground, the person needs to go as fast as approximately 5.96 m/s

Explanation:

The height of the hot air balloon above the ground, s = 92.4 m

The location on the ground of the person intending to catch the box, x = 10 m away horizontally from the base location of the balloon

The time it takes the balloon released at zero vertical velocity is given by the following equation;

s = 1/2·g·t²

92.4 = 1/2 × 9.8 × t² = 19.6·t²

t² = 92.4/19.6 ≈ 4.714

t = √4.714 ≈ 1.68 seconds

The time, t, it takes the box to hit the ground ≈ 1.68 seconds

The speed, v, of the person standing 10 m away has to go to catch the box before it hits the ground is v = Distance/time =10/1.68 ≈ 5.96 m/s

In order to catch the box before it hits the ground, the person needs to go as fast as approximately 5.96 m/s.

which statement about this sum is true? −245+2.7

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