use v (t ) = A (1 − e − kt^ ). Generally, the k term is going to be whatever’s multiplying the v in the original
equation. In this case, then, k = b/m .
Now, look at the initial and final conditions to find the value of A . At time t = 0, v = 0; that works no
matter the value of A , because e 0 = 1. But after a long time, we know the terminal velocity is mg /b . And
in the equation, e −kt goes to zero for large t . Meaning: after a long time, the velocity function equals A .
This A must be the terminal velocity!
So our final equation looks like: .Practice Problems
Multiple Choice:
1 . A firework is shot straight up in the air with an initial speed of 50 m/s. What is the maximum height it
reaches?
(A) 12.5 m
(B) 25 m
(C) 125 m
(D) 250 m
(E) 1250 m2 . On a strange, airless planet, a ball is thrown downward from a height of 17 m. The ball initially
travels at 15 m/s. If the ball hits the ground in 1 s, what is this planet’s gravitational acceleration?
(A) 2 m/s^2
(B) 4 m/s^2
(C) 6 m/s^2
(D) 8 m/s^2
(E) 10 m/s^23 . An object moves such that its position is given by the function x (t ) = 3t 2 − 4t + 1. The units of t are
seconds, and the units of x are meters. After 6 s, how fast and in what direction is this object moving?
(A) 32 m/s in the original direction of motion
(B) 16 m/s in the original direction of motion
(C) 0 m/s
(D) 16 m/s opposite the original direction of motion
(E) 32 m/s opposite the original direction of motionFree Response:
4 . An airplane attempts to drop a bomb on a target. When the bomb is released, the plane is flying
upward at an angle of 30° above the horizontal at a speed of 200 m/s, as shown below. At the point of
release, the plane’s altitude is 2.0 km. The bomb hits the target.