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?

The person has to move at the speed of 2.30 m/s towards the dropping point to catch the box.

Explanation:

Let t be the time taken by the box to reach the ground and v be the speed of the person on the ground standing 10 m away from the dropping

point.

As distance=(speed) x (time), so

Initially, the height of the box from the ground is 92.4 m.

So, the displacement, s, covered by the box when it reached the ground,

m in the downward direction.

The gravitational force action of the box is also in the downward direction.

The initial velocity of the balloon, u=0.

According to the equation of motion,

where a is the acceleration, u is initial velocity and.

Here, (acceleration due to gravity) and s and a are having the same direction.

So, on putting the values in the equation, we have

seconds.

So, the time available for the man on the ground to catch the bot is 4.34 seconds.

Now, from equation (i),

m/s.

Hence, the person has to move at the speed of 2.30 m/s towards the dropping point to catch the box.

i will answer your question if you answer mine

it was given as a lesson and guide for prayer.