A plane drops a package of food to stranded caribou on an ice flow in the Arctic circle.The plane flies 274.47 m above the ice at a speed of 162.49 m/s. How far short of the target should it drop...

A plane drops a package of food to stranded caribou on an ice flow in the Arctic circle.

The plane flies 274.47 m above the ice at a speed of 162.49 m/s. How far short of the target should it drop the package?

Asked on by przybyh

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Given an initial height of 274.47m and an initial velocity of 162.49m/s, find the range of the package. We assume that we can ignore air resistance.

Here are two approaches:

(1) The package follows a path along a parabola, but the time it takes to hit the ground is the same regardless of the horizontal motion. Using the falling body formula `h=-4.9t^2+v_0t+s_0` where h is the height at time t (h in meters, t in seconds), `v_0` is the initial vertical velocity and `s_0` the initial height we get:

`-4.9t^2+274.47=0==>t=7.484` so the time it takes for the package to fall to the ground is approximately 7.484 seconds.

The horizontal motion is separate from the vertical motion, so if the package is traveling at a velocity of 162.49m/s, in 7.484 seconds it will have travelled 162.49*7.4841=1216.075m.

Thus the plane should drop the package 1216.08m from the target.

(2) The distance (d) travelled by a projectile (ignoring air resistance) from a given height (`y_0`) and at a given velocity (`v`) at a given angle `theta` is given by:

`d=(vcos theta)/(g) (vsin theta + sqrt((vsin theta)^2+2gy_0)) `

Since `theta=0` we have `cos theta=1,sin theta=0` and with `g=9.8,y_0=274.47,v=162.49` we get:

`d=162.49/9.8(sqrt(2(9.8)(274.47))`

`d~~1216.119`

So again we see that the package should be dropped approximately 1216.1m from the target.

Sources:

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