A rescue plane spots a person floating in a lifeboat 55 m directly below and releases an emergency kit with a parachute. The package descends with a constant vertical acceleration of 6.91 m/s. If the horizontal speed of the plane was 70.6 m/s, how far is the package from the lifeboat when it hits the waves?

Question

A rescue plane spots a person floating in a lifeboat 55 m directly below and releases an emergency kit with a parachute. The package descends with a constant vertical acceleration of 6.91 m/s. If the horizontal speed of the plane was 70.6 m/s, how far is the package from the lifeboat when it hits the waves?

225 m

1.27 km

323 m

282 m

Answer 1

Would it be 282 m?

Answer 2

vertical acceleration of 6.91 m/s>>> I bet the writer mean vertical velocity.

calcualte time in air; height/speedabove
then, horizontal distance: time in air*initial horizontal veloicty

Answer 3

Thank you!

Answer 4

To find the distance between the package and the lifeboat when it hits the waves, we can first calculate the time it takes for the package to reach the lifeboat's height.

We can use the equation of motion for vertical motion:

v = u + at

Where:
v = final velocity (0 m/s, as the package reaches the lifeboat's height)
u = initial velocity (0 m/s, as the package is released from rest)
a = acceleration (-6.91 m/s^2, as the package is decelerating due to gravity)
t = time taken

Rearranging the equation, we have:

t = (v - u) / a
t = (0 - 0) / -6.91
t = 0 / -6.91
t = 0 seconds

As we can see, it takes 0 seconds for the package to reach the lifeboat's height. This means that the package immediately starts falling vertically.

Now, let's find the time it takes for the package to reach the waves. We can use the equation:

s = ut + (1/2)at^2

Where:
s = distance traveled (unknown)
u = initial velocity (0 m/s, as the package is released from rest)
a = acceleration (-9.8 m/s^2, as the package is accelerating due to gravity)
t = time taken (unknown)

Rearranging the equation, we have:

s = (1/2)at^2
s = (1/2)(-9.8)t^2
s = -4.9t^2

Substituting t = 0 into the equation, we get:

s = -4.9(0)^2
s = 0 meters

This means that the package does not initially have any horizontal distance from the lifeboat. Therefore, the correct answer is 0 meters. However, none of the given options match this result.

Answer 5

55/6.91 = 7.96

7.96 * 70.6 = 562

That is not one of the listed answers. I suspect using acceleration of 6.91 m/s^2 works better.

Answer 6

Somehow my answer did not appear

55 = (1/2) 6.91 t^2
t = about 4
4 * 71 = 284
yes, 282