A helicopter is flying in a straight line over a level field at a constant speed

Question

A helicopter is flying in a straight line over a level field at a constant speed of 5.4 m/s and at a constant altitude of 8.9 m. A package is ejected horizontally from the helicopter with an initial velocity of 15.0 m/s relative to the helicopter, and in a direction opposite the helicopter's motion. The initial velocity of the package relative to the ground is -9.6m/s. The angle the velocity vector of the package makes with the ground at the instant before impact, as seen from the ground is 54 deg.


How would you calculate the angle the velocity vector makes with the ground at the instant before impact, as seen from the ground?  ( know the answer is 54 degrees, but I want to know how they derived it).  Thanks!

Explanation / Answer

in y -direction

y = u*t + 0.5*g*t^2

9 = 0 + 0.5*9.8*t^2

9 =4.9*t^2

t = sqrt(9/4.9) = 1.414 s

vy = u + g*t = 0 +9.8*1.414 = -13.86 m/s (- sign indicates down ward)

vx = -9.6 m/s

theta = tan^-1(vy/vx) = tan^-1(13.86/9.6) = 54 degrres with horizontal

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