Question

An airplane with a speed of 85.3 m/s is climbing upward at an angle of 45.0 ° with respect to the horizontal. When the plane's altitude is 587 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

what I'm doing is that Im finding voy=sin45*85.3=60.32
then i'm finding the time by using the equation vot+0.5at^2
then i'm using the equation D=v*t to find the distance, but i'm getting the wrong answer. pls help clarify the steps and what i'm doing wrong.

Answer 1

Vi = 60.32 yes

u = 85.3 cos 45 = 60.32 also but forever

v = Vi + a t = 60.32 -9.81 t
so at top when v = 0
t = 6.15 seconds coasting up

h at top = Hi + Vi t -4.9 t^2
= 587 + 60.32 (6.15) - 4.9(6.15^2)
= 773 meters high at 6.15 seconds when fall starts

now it falls
773 = 4.9 t^2
t = 12.6 second fall
so
total time after relaese = 12.6+6.15 = 18.7 seconds at horizontal velocity 60.32
d = 18.7 * 60.32 = 1,128 meters

velocity
u = 60.32 we know
v = g t = 9.81*12.6 = 124 down, negative really

speed = sqrt (u^2+v^2) = 138 m/s
angle down from horizontal = A
tan A = v/u = 124/60.32
A = 64 degrees below horizontal

Answer 2

thank you very much Damon. It is much clear now, since u explained it step by step.