An airplane with a speed of 76.8 m/s is climbing upward at an angle of 48° with respect to the horizontal. When the plane's altitude is 560 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
X-component of the package velocity on the instant when it begins the motion is
vx= vcosα=76.8cos48^0=51.4m/s.
The description of this projectile motion is
x=vxt
y=(gt^2)/2,
the time of motion is t=sqroot(2y/g)=10.7 s,
then the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth, is
x=51.4•10.7=549.5 m.
tanα= vy /vx =gt/ vx=9.8•10.7/51.4=0.324
α=arctan 0.324=18 deg.