After discovering that two hikers have been stranded on Devil\'s Thumb Pass, the

Question

After discovering that two hikers have been stranded on Devil's Thumb Pass, the Colorado Search and

Rescue team sends out a plane with supplies for the hikers. Because the hikers are in an isolated location, the

plane has to drop the supplies as it ies overhead. If the plane is 235 m above the hikers and

has a horizontal speed of 200 km/h, how far away from the hikers should the supplies be dropped if they are

to land at the hikers' location? (Hint: the initial y-component of the supplies' velocity is zero

Explanation / Answer

Solution: We consider the horizontal direction as X direction and vertical direction as Y direction. Further we consider the origin at the plane the moment it drops the supplies.

Thus the plane’s height above the hiker h = 235 m

Horizontal speed of the supplies when they were dropped; v = 200 km/h = (200 km/hr)*(1000m/1km)/(3600s/1hr) = 55.56 m/s

The angle made by the supplies when were being dropped = 0o

The horizontal component of the velocity vx = v*cos = 55.56*cos(0o) = 55.56 m/s

The vertical component of the velocity vy = v*sin = 55.56*sin(0o) = 0 m/s

Its means that the vertical velocity of the supplies is zero as they were just dropped (not thrown) from a horizontally moving plane. However the supplies have the horizontal velocity equal to 55.56 m/s initially.

When the supplies reach victims, the vertical displacement of the supplies is y = -235 m. (Negative sign taken for the distance fallen by the supplies in the negative Y direction)

The time taken by the supplies to fall this vertical distance is given by,

y = vy*t +(1/2)*g*t2

-235m = (0m/s)t + (1/2)*(-9.81)* t2

t2 = 47.91 s2

t = 6.9217 s.

Thus the supplies fall 235 m in 6.9217 seconds.

During this fall, the supplies also move horizontally with the velocity of vx = 55.56 m/s.

(The horizontal component of velocity do not change as there is no acceleration acting in the horizontal direction)

Thus the horizontal distance travelled by the supplies in t = 6.9217 s is given by

x = vx*t

x = (55.56 m/s)*(6.9217 s)

x = 384.57 m.

Thus the supplies should be dropped 384.57 m or 385 m away from the hiker (before flying overhead) if they to land at the hiker’s location.

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