Question

An airplane with a speed of 92.0 m/s is climbing upward at an angle of 41.9 ° with respect to the horizontal. When the plane's altitude is 745 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.

Answer 1

a)

Considering motion along vertical

h = ho + u sin x * t - 0.5 gt^2

0 = 745 + 92* sin 41.9 * t - 4.9 t^2

solving for t,

t = 20.1 seconds

Considering motion along horizontal

R = u cos x * t = 92* cos 41.9 * 20.1

R = 1376. 38 m

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b)

vertical velocity,

vy = - u sin x + gt

vy = - 92 sin 41.9 + 9.8* 20.1

vy = 135.54 m/s

Horizontal velocity

vx = 92 cos 41.9 = 68.48 m/s

angle of velocity

phi = arctan ( Vy/ vx)

phi = 63.195

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