1. A DC-8 flying horizontally at an altitude of 20km at a speed of 250m/s
drops an engine.
2. An arrow is fired at 53° with a velocity of 15m/s.
a. How long after it falls does it take the engine to hit the ground?
b. How far horizontally has the engine traveled when it hits?
a. 63.888s
b. 15972m
For part a, we need to simply use the distance substitution equation.

-20,000m = 0m/s * t + 0.5 * -9.8m/s2 * t^2

-20,000m = -4.9m/s2 * t^2

4081.633s2 = t^2

63.888s = t
For part b, we need only multiply the speed by the time.

250m/s * 63.888s = 15972m
a. How long is it in the air?
b. What is its range?
a. 2.445s
b. 22.071m
First, let's find the horizontal and vertical components. Then we'll
find part a using the acceleration equation.

15m/s cos 53° = 9.027m/s

15m/s sin 53° = 11.979m/s
-9.8m/s2 = (-11.979m/s - 11.979m/s) / t

2.445s = t
To find the range, all we have to do is multiply the time by the
horizontal velocity.

2.445s * 9.027m/s = 22.071m

