Answer:
The person has to move at the speed of 2.30 m/s towards the dropping point to catch the box.
Step-by-step 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.