If the plane's speed is v, then
the package's initial downward speed is v sin(-34°)
So, if it takes 6.4s to land, then
490 - v sin34° * 6.4 - 4.9*6.4^2 = 0
v = 80.84 m/s
The horizontal speed is 80.84 cos34° = 67 m/s
distance = speed * time = 67 * 6.4 = 428.8 m
A plane dives at 34∘ to the horizontal and releases a package at an altitude of 490 m. If the load is in the air for 6.4 s, find:
a) the speed of the plane when it released the package;
b) the horizontal distance traveled by the package after it is released.
1 answer