Projectile & Two-Dimensional Motion — Practice · Grade 12 Science
Chapter: Motion & KinematicsWhat this practice covers
By the end of Projectile & Two-Dimensional Motion you should be able to:
- I can explain that the horizontal and vertical motions of a projectile are independent.
- I can analyze horizontal motion as constant velocity and vertical motion as free fall under g.
- I can solve for a projectile's time of flight, range, and landing position.
10 questions, numeric, multiple choice and written answer, each with a worked explanation.
Vocabulary used: projectile, trajectory, horizontal component, vertical component, range, independence of motion.
Problem 1
Answer and explanation
2 m
Horizontal motion is constant velocity: x = (4)(0.5) = 2 m.
Problem 2
- decreases to zero
- depends on the object's weight
- increases as it falls
- stays constant
Answer and explanation
stays constant
With no horizontal force acting, Newton's first law keeps the horizontal velocity constant for the whole flight.
Problem 3
Answer and explanation
3 s
Vertical motion: 44.1 = ½(9.8)t² = 4.9t², so t² = 9.0 and t = 3.0 s.
Problem 4
- the fired bullet lands first
- the dropped bullet lands first
- it depends on the bullet's speed
- both land at the same time
Answer and explanation
both land at the same time
Vertical motion is independent of horizontal motion, so both bullets are in identical free fall and land together.
Problem 5
Answer and explanation
80 m
First find the time: 78.4 = 4.9t², so t² = 16 and t = 4.0 s. Then range = (20)(4.0) = 80 m.
Problem 6
Answer and explanation
Once both are in the air, the dart and the monkey fall with the same downward acceleration g. In the time the dart travels to the monkey's original spot, gravity pulls the dart below its straight-line aim by exactly ½gt² — the same distance the monkey has fallen in that time. Because both drop by the identical amount, they meet. The dart's speed only changes when they meet, not whether they meet.
Problem 7
Hints
Solve vertical motion for time before using horizontal speed.
Answer and explanation
45=5t² gives t=3 s; range=18 m.
Problem 8
Hints
State what the supplied observations or conditions establish, and separate that from any assumption.
Answer and explanation
No. Vertical velocity is momentarily zero, but horizontal velocity remains nonzero without drag.
Problem 9
Answer and explanation
No. The vertical equation is unchanged; range doubles while the same flight time applies.
The two components have separate equations but share the same elapsed time. Once the vertical equation gives the later launch-level landing at about 3.27 s, horizontal displacement is 12×3.27≈39.2 m. Recombining the landing components gives speed √(12²+16²)=20 m/s. This equality with launch speed follows equal height and negligible drag; it is not a universal rule for every landing elevation or real ball.
No. The vertical equation is unchanged; range doubles while the same flight time applies.
Response criteria
- Interpret labels, structures, axes, units and directions correctly.
- Explain the physical or environmental relationship or mechanism.
- Respect stated assumptions and distinguish observation from inference.
Problem 10
Answer and explanation
0=19.6−4.9t² gives t=2.00 s. Range=8×2=16.0 m. Landing velocity=(8.0,−19.6) m/s in right/up-positive axes; speed≈21.2 m/s.
0=19.6−4.9t² gives t=2.00 s.
Range=8×2=16.0 m.
Landing velocity=(8.0,−19.6) m/s in right/up-positive axes; speed≈21.2 m/s.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.