a) The box must be dropped 14.3 seconds before
b) The horizontal distance between the plane and the victims is 496.2 m
Explanation:
a)
The motion of thebox is a projectile motion, therefore it consists of two independent motions:
- A uniform motion (at constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (=acceleration of gravity, [tex]g=9.8 m/s^2[/tex]) in the downward direction
To solve part a) of the problem, we just need to analyze the vertical motion. We can do it by using the following suvat equation
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s = 1000 m is the vertical distance that the box must cover
u = 0 is the initial vertical velocity of the box (zero, since it is travelling horizontally)
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
t is the time
Solving for t, we find the time the box need to reaches the ground:
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(1000)}{9.8}}=14.3 s[/tex]
So, the box must be dropped 14.3 s before.
b)
The airplane is travelling with a uniform motion along the horizontal direction, therefore we can use the following equation:
[tex]d=vt[/tex]
where
d is the horizontal distance travelled
v is the horizontal velocity
t is the time elapsed
For this airplane, we have:
[tex]v=125 km/h \cdot \frac{1000 m/km}{3600 s/h}=34.7 m/s[/tex] is the velocity
t = 14.3 s is the total time of flight of the box
Solving for d, we find the horizontal distance between the plane and the victims when the box is dropped:
[tex]d=(34.7)(14.3)=496.2 m[/tex]
Learn more about projectile motion:
brainly.com/question/8751410
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