An airplane is flying with a velocity of 81.0m/s at an angle of 21.0∘ above the horizontal.?
An airplane is flying with a velocity of 81.0m/s at an angle of 21.0∘ above the horizontal. When the plane is a distance 112m directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment.
How far from the dog will the suitcase land? You can ignore air resistance.
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Height at any time t = y(t) = h + Uy t - 1/2 g t^2
Distance at any time t = x(t) = Ux t
g = 9.81 m/s^2
Launch height H = 112.000 meters above ground
Impact elevation y(T) = 0.000 meters
h = H - y(T) = 112.000 meters re impact elevation
Launch speed U = 81 mps 291.600 kph
Launch angle theta = 21 degrees re the horizontal
Uy = U sin(theta) = 29.028 mps 104.500 kph
Ux = U cos(theta) = 75.620 mps 272.232 kph
Solve the quadratic: 0 = 112.000 + 29.028 T - 4.905 T^2 ... for T
A = 4.905
B = -29.028 Quadratic coefficients
C = -112.000
Total Flight Time to impact elevation(s) sec
T = 8.579
T' = -2.661
Range(s) to impact elevation(s) meters
x(T) = Ux T = 75.620 * 8.579 = 648.779 <=== ANS, assuming the dog is asleep
x(T) = Ux T = 75.620 * -2.661 = #N/A
NOTE: There could be two ranges, the early and the later
Time to reach max height sec
tmax = 2.959
Range at Max Height meters
x(tmax) = Ux tmax = 223.760
Max Height above impact elevation meters
y(max) = 154.947
Max Height above h meters
y(max) - h = 42.947
Vy at impact mps
Vy = Uy - gT = 29.028 - 9.810 * 8.579 = -55.137
Vx at impact mps
Vx = Ux = 75.620
Total impact speed mps
V = sqrt(Vy^2 + Vx^2) = 93.587
Angle of impact re horizontal degrees
psi = ATAN(Vy/Vx) = ATAN(-55.137/75.620) = -36.097
HERE's another question. At max height for the suitcase, will the aircraft be clear of that suitcase by continuing its climb rate when the suitcase is dropped?