An airplane with a speed of 92.9 m/s is climbing upward at an angle of 32.5° wit

Question

An airplane with a speed of 92.9 m/s is climbing upward at an angle of 32.5° with respect to the horizontal. When the plane's altitude is 612m, 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. the answer to a.) is 1360.37547 I am stuck on part b.) and cannot get the angle correct. any help would be great thanks!

Explanation / Answer

Here,
initial speed , u = 92.9 m/s

theta = 32.5 degree

h = 612 m

a)

let the time taken for the package to fall is t

Using second equation of motion

-612 = 92.9 * sin(32.5 degree) * t - 0.5 * 9.8 * t^2

solving for t

t = 17.4 s

total distance from the ground = 17.4 * 92.9 * cos(32.5 degre)

total distance from the ground = 1363 m

b) for the final vertical velocity

vy = 92.9 * sin(32.5 degree) - 9.8 * 17.4 = -121.5 m/s

vy = 92.9 * cos(32.5 degre) = 78.5 m/s

angle just before impact = arctan(vy/vx)

angle just before impact = arctan(121.5/78.5) below horizontal

angle just before impact = 57.2 degree below horizontal

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