Projectile Motion

Two independent motions, happening at once.

12 problems · about 22 minutes · TEKS P.3.C

What this lesson teaches

The student models projectile motion parametrically and finds range, maximum height and flight time.

Horizontal and vertical motion are independent of each other.
x = (v cos θ)t and y = (v sin θ)t − 4.9t².
The peak occurs halfway through the flight, and 45° maximises range on level ground.

Straightforward practice. Get the method working first.

5 problems

Launch at 30 m/s at 30°. What is the vertical component, in m/s?
Answer 15
Why 15 m/s.
What happens to the horizontal speed during flight, ignoring air resistance?
Answer It stays constant, because gravity acts only downward.
Why It stays constant.
Launch at 40 m/s at 0°. What is the vertical component, in m/s?
Answer 0
Why sin 0° = 0.
Launch at 40 m/s at 90°. What is the horizontal component, in m/s?
Answer 0
Why cos 90° = 0.
Launch at 10 m/s at 30°. What is the vertical component, in m/s?
Answer 5
Why 10 sin 30°.

The same ideas with more to keep track of.

3 problems

Flight time 6 seconds. When is the peak, in seconds?
Answer 3
Why Halfway.
Horizontal speed 15 m/s for 4 seconds. What is the range, in metres?
Answer 60
Why Speed times time.
Which launch angle gives the greatest range on level ground, in degrees?
Answer 45
Why It splits the speed evenly.

Mixed problems. Work out what kind of question it is before you start.

4 problems

Sort each quantity by which motion it belongs to.
Answer Horizontal: v cos θ, The range · Vertical: v sin θ, The maximum height
Why Cosine goes across; sine goes up.
Horizontal speed 20 m/s for 3 seconds. Range in metres?
Answer 60
Why 20 × 3.
The Components: A ball is launched at 20 m/s at 30°. What is the vertical component, in m/s?
Answer 10 m/s
Why 10 m/s.
The Halfway Point: A projectile is in the air for 4 seconds. At what time, in seconds, does it reach its highest point?
Answer 2 seconds
Why 2 seconds.