An airplane traveling 1001 m. above the ocean at 125 km/h is going to drop a box of supplies to shipwrecked..?

Question

An airplane traveling 1001 m. above the ocean at 125 km/h is going to drop a box of supplies to shipwrecked victims below.

a. how many seconds before the plane is directly overhead show the box be dropped?

b. What is the horizontal distance between the plane and the victims when the box is dropped?

This problem is giving me trouble.

I don't understand part a, and even more so part b.
Please help.

Answer 1

ok

To start use equation d(sub x sub f) =d( sub x sub i) + V(sub 0 sub x) t + 1/2 at^2

Say d(sub x sub i) is where the package starts so it is zero. There is no acceleation of the box in the x direction so it's just vt. v is the speed of the plane, make sure you change it from 125 km/hour to m/s. do 125 times 1000, then divide by 3600.

just leave it alone for now

Now, same equation for the y-direction.

d( sub y sub f) = d(Sub y sub i) + v( sub y sub i) t + 1/2 at^2

Now, in this equation, down is positive, so d(sub i) = 0

the distance you plug in the height of the plane for that., v (sub i) is 0, a is 9.8.

solve for t. you should get a number, thats answer to a

Plug that number for t into the first equation for t. the d (sub x) equation, and you should get the distance that the package travels after it is dropped

Answer 2

time for the box to drop in the water: t=sqr(2d/g)
t=sqr(2*1001m/9.8)=14.29s
125km/hr--->34.72m/s
x=vt
x=(34.72m/s)(14.29s)
x=496.18m
The plane has to drop the box 496.18m away from the victim.

Answer 3

a. d_v = (1/2)at^2
t = sqrt(2d_v/a) = sqrt(2(1001)/9.81) = 14.3 s

b. d_h = v_h(t) = 125km/h(1000m)/(3600s)t = 496 m

Answer 4

Makes two of us.