Question

A plane traveling horizontally at 120 ​m/s over flat ground at an elevation of 3610 m must drop an emergency packet on a target on the ground. The trajectory of the packet is given by xequals 120​t, yequals minus4.9tsquaredplus3610​, for tgreater than or equals ​0, where the origin is the point on the ground directly beneath the plane at the moment of the release. How many horizontal meters before the target should the packet be released in order to hit the​ target? Set up an equation involving t that could be solved in order to find out how long it takes the packet to reach the ground after being released.

Answer 1

Answer:

$x=3257.143\ m$

Explanation:

Given:

• horizontal velocity of the plane,

• elevation of the plane,

Trajectory of the packet is given by:

$x=120.t$ ...........................(1)

$y=-4.9.t^2+3610$ ...................(2)

According to given at ground level we have y = 0.

Putting this condition in eq. (2)

$0=-4.9.t^2+3610$

$t=27.14\ s$

Now the horizontal distance travelled by the plane in the above time:

(Due to inertia of motion the packet will also travel the same distance horizontally from the point where it is dropped.)

$x=120\times 27.14$

$x=3257.143\ m$ is the horizontal distance from the target on the earth before which the packet must be released in order to reach there.

Answer 2

Answer:

Explanation:

Horizontal displacement

x = 120 t

Vertical position

y = 3610 - 4.9 t²

y = 0 for the ground

0 = 3610 - 4.9 t²

t = 27.14 s

This is the time it will take to reach the ground .

During this period , horizontal displacement

x = 120 x 27.14 m

= 3256.8 m

So packet should be released 3256.8 m before the target.