An airplane with a speed of 34 m/s is climbing upward at an angle of 46° with respect to the horizontal...?
An airplane with a speed of 34 m/s is climbing upward at an angle of 46° with respect to the horizontal. When the plane's altitude is 780 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.
__________ m
(b) Relative to the ground determine the angle of the velocity vector of the package just before impact.
________ ° clockwise from the positive x axis
Answers & Comments
Verified answer
34sin 46 = 24.46 = Vy
34cos46 = 23.62 = Vx
x = x0 +v0t +1/2At^2
x = 780 +24.46t - 4.9t^2 = 0
4.9t^2 -24.46t -780 = 0
quad formula
24.46/9.8 +/-sqrt(598.29+38220)/9.8
24.46+/-28.30 = -3.839 or 52.759
52.759 sec to reach ground
23.62m/s = Vx
Vx *52.759 sec = distnce from drop = 1246meters