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  1. A plane flying horizontally at an altitude of 490 meters and having a velocity of 250 m/s East, drops a supply packet to a work crew on the ground.  It falls freely without a parachute.  Assume no wind and no air resistance.
    1. Sketch a path of the supply packet after it leaves the plane
  2. Determine the time required for the packet to hit the ground. (Make lists)
in Grade 12 Physical Sciences by Diamond (39.8k points) | 171 views

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If you picture this scenario in your mind then the motion of package is that of a projectile. For solving this Question you have to know that the horizontal velocity component of a body undergoing projectile motion is constant.

And vertical component of its velocity changes with an acceleration of 9.8m/s(g).

Now, when the parcel is dropped its initial velocity is zero in vertical direction and its initial velocity is 259 m/s in horizontal direction which will remain as it is.

Since, height from which the parcel is dropped is 980m. And time taken to cover distance with zero initial velocity and 9.8m/s of acceleration is is proportional to distance from which it falls.

Putting above values in equation:

S = ut + 0.5(a)t^2

490 = 0 + 0.5(9.8)(t)^2

490 = 4.9(t)^2

100 =(t)^2

t=root(100)

t=10s

by Diamond (39.8k points)
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1.

 

2. Assume right = + ; left = - ; down = - ; up = +

yf = final height of packag

yi = initial height of package

Vi = initial velocity

a / g = acceleration ( down, so we'll use - sign)

t = time

yf = yi + (Viy)(t) + 1/2gt²

0 = 490 + (0)t + 1/2(-9.8)t²

0 = 490 - 4.9t²

t² = 490 / 4.9

t = √490 / 4.9

t = 10 secs

Additionally (for interest sake):

To get the distance of the package travelled = x:

X = (Vix)(t)            Vix = here the initial velocity of the package is the same as the plane

    = (250)(10)

    = 2500m

To find the angle ϴ:

tanϴ = opposite / adjacent

tanϴ = 490 / 2500

ϴ = tan^-1 (490/2500)

ϴ = 11.09°

 

by Wooden (734 points)
0 0

Another diagram. 

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