Peoples Physics Concepts

http://www.ck12.org Chapter 3. Two-Dimensional and Projectile Motion

Guidance

• To work these problems, separate the “Big Three” equations into two sets: one for the vertical direction, and
  one for the horizontal. Keep them separate.


• The only variable that can go into both sets of equations is time; use time to communicate between the x and
  y components of the object’s motion.

Example 1

CSI discovers a car at the bottom of a 72 m cliff. How fast was the car going if it landed 22m horizontally from the
cliff’s edge? (Note that the cliff is flat, i.e. the car came off the cliff horizontally).

Question: v=?[m/s]

Given: h=∆y= 72 m

d=∆x= 22 m

g= 10. 0 m/s^2

Equation: h=viyt+^12 gt^2 andd=vixt

Plug n’ Chug:Step 1: Calculate the time required for the car to freefall from a height of 72 m.

h=viyt+^12 gt^2 but sinceviy=0, the equation simplifies toh=^12 gt^2 rearranging for the unknown variable,t, yields

t=

√

2 h

g

=

√

2 ( 72 m)

10. 0 m/s^2

= 3. 79 s

Step 2: Solve for initial velocity:

vix=dt = 322. 79 ms= 5. 80 m/s

Answer:

5 .80 m/s

Example 2

Question: A ball of massmis moving horizontally with a speed ofvioff a cliff of heighth. How much time does it
take the ball to travel from the edge of the cliff to the ground? Express your answer in terms ofg(acceleration due
to gravity) andh(height of the cliff).

Solution: Since we are solving or how long it takes for the ball to reach ground, any motion in thexdirection is not
pertinent. To make this problem a little simpler, we will define down as the positive direction and the top of the cliff
to be

y= 0

. In this solution we will use the equation

y(t) =yo+voyt+

1

2

gt^2

.