Answer:
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
[tex]10=vt\cdots(i)[/tex]
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,
[tex]s=92.4[/tex] 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,
[tex]s=ut+\frac1 2 at^2[/tex]
where a is the acceleration, u is initial velocity and.
Here, [tex]a=g=9.81 m/s^2[/tex] (acceleration due to gravity) and s and a are having the same direction.
So, on putting the values in the equation, we have
[tex]92.4=0\timest+ \frac 1 2 (9.81)t^2[/tex]
[tex]\Rightarrow t^2=\frac{92.4\times 2}{9.81}[/tex]
[tex]\Rightarrow t=\sqrt{\frac{92.4\times 2}{9.81}}[/tex]
[tex]\Rightarrow t=4.34[/tex] seconds.
So, the time available for the man on the ground to catch the bot is 4.34 seconds.
Now, from equation (i),
[tex]10=v\times 4.34[/tex]
[tex]\Rightarrow v = 10/4.34[/tex]
[tex]\Rightarrow v = 2.30[/tex] m/s.
Hence, the person has to move at the speed of 2.30 m/s towards the dropping point to catch the box.
