Projectile & Two-Dimensional Motion — Practice · Grade 12 Science

Chapter: Motion & Kinematics

What this practice covers

By the end of Projectile & Two-Dimensional Motion you should be able to:

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

A ball rolls off a table and is in the air for 0.5 s. It left the table moving horizontally at 4 m/s. How far from the base of the table does it land?
Answer and explanation

2 m

Horizontal motion is constant velocity: x = (4)(0.5) = 2 m.

Problem 2

Ignoring air resistance, the horizontal velocity of a projectile during its flight…
  1. decreases to zero
  2. depends on the object's weight
  3. increases as it falls
  4. 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

A stone is thrown horizontally from a bridge 44.1 m above the water. How long until it hits the water? (g = 9.8 m/s²)
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

A bullet fired horizontally and an identical bullet simply dropped from the same height at the same instant will…
  1. the fired bullet lands first
  2. the dropped bullet lands first
  3. it depends on the bullet's speed
  4. 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

A ball is launched horizontally at 20 m/s from a cliff 78.4 m high. How far from the base of the cliff does it land? (g = 9.8 m/s²)
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

In the classic 'monkey and hunter' demonstration, a monkey drops from a branch at the exact instant a dart is fired straight at it. Explain why the dart always hits the monkey, no matter the dart's speed.
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

From 45 m height with horizontal speed 6 m/s and g=10, find flight time and range. Show the reasoning, not just the conclusion.
Hints

Solve vertical motion for time before using horizontal speed.

Answer and explanation

45=5t² gives t=3 s; range=18 m.

Problem 8

At the top of a nonvertical projectile path, is total velocity zero?
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

Use the same flight time in both perpendicular componentsUse the same flight time in both perpendicular componentsStepRelationshipResultFind landing timet=2vy₀/g3.27 sHorizontal travelx=vxt39.2 mLanding velocity(12,−16) m/sSpeed 20 m/sA shared clock connects the two component models.
Original constant-g, no-drag calculation; range and speed are model results, not measured throws.
Interpret the supplied visual. Would doubling vx while retaining vy₀ change the modeled launch-level flight time?
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

A ball is launched horizontally at 8.0 m/s from a 19.6 m ledge. Ignore drag and use g=9.8 m/s². Find flight time, horizontal range and landing velocity components.
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.

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