A relief airplane is delivering a food package to a group of people stranded on a very small island. The island is too small for the plane to land on, and the only way to deliver the package is by dropping it. The airplane flies horizontally with constant speed of 200 at an altitude of 800 . The positive x and y directions are defined in the figure. For all parts, assume that the "island" refers to the point at a distance from the point at which the package is released, as shown in the figure. Ignore the height of this point above sea level. Assume that the acceleration due to gravity is = 9.80 .
After a package is ejected from the plane, how long will it take for it to reach sea level from the time it is ejected? Assume that the package, like the plane, has an initial velocity of 200 in the horizontal direction.
t=12.8
If the package is to land right on the island, at what horizontal distance from the plane to the island should the package be released?
D=1140m
I could not answer What is the speed of the package when it hits the ground?
Express your answer numerically in miles per hour.?
I got 281 mph but it was wrong :( please help
After a package is ejected from the plane, how long will it take for it to reach sea level from the time it is ejected? Assume that the package, like the plane, has an initial velocity of 200 in the horizontal direction.
t=12.8
If the package is to land right on the island, at what horizontal distance from the plane to the island should the package be released?
D=1140m
I could not answer What is the speed of the package when it hits the ground?
Express your answer numerically in miles per hour.?
I got 281 mph but it was wrong :( please help
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From h = 800 m, the time to impact is T = sqrt(2h/g) = sqrt(2*800/9.8) = 12.78 sec. ANS.
In that time, the package will travel X = Ux T = (200/3600)*12.78 = 0.71 miles; so the release point has to be .71 mi ~ 1142.63 m shy of X marks the spot on the island. ANS.
V = sqrt(Vy^2 + Ux^2); where Vy^2 = 2gh, so V = sqrt(2*9.8*800 + 89.4^2) = 153.86 mps = 344.18 mph at an angle theta = ATAN(-sqrt(2*9.8*800)/89.4) = -54.47 degree below the horizontal. ANS. Note Ux = 200 mph = 89.4 mps.
In that time, the package will travel X = Ux T = (200/3600)*12.78 = 0.71 miles; so the release point has to be .71 mi ~ 1142.63 m shy of X marks the spot on the island. ANS.
V = sqrt(Vy^2 + Ux^2); where Vy^2 = 2gh, so V = sqrt(2*9.8*800 + 89.4^2) = 153.86 mps = 344.18 mph at an angle theta = ATAN(-sqrt(2*9.8*800)/89.4) = -54.47 degree below the horizontal. ANS. Note Ux = 200 mph = 89.4 mps.
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put some units in with your measurements...