Projectile Motion at an Angle- Practice Questions and Solutions

Projectile Motion at an Angle: What You Actually Need to Know

Most students screw up projectile motion problems because they treat horizontal and vertical motion as the same thing. They're not. This guide cuts through the confusion with practice questions and clear solutions.

The Core Concept

An object launched at an angle follows a curved path called a parabola. Gravity only pulls it down. Nothing pushes it forward. The horizontal velocity stays constant because there's no horizontal force acting on it.

That's the whole deal. Now let's break it down.

The Two Components

• Horizontal motion — velocity is constant (vx = v₀ cos θ). Air resistance is ignored.
• Vertical motion — velocity changes due to gravity (vy = v₀ sin θ - gt). Initial velocity going up, acceleration going down.

Key Formulas

g = 9.8 m/s² (or 10 m/s² for simplicity). θ = launch angle. v₀ = initial speed.

Quick Reference: What Changes and What Doesn't

Practice Questions

Question 1: Basic Launch

A ball is thrown with an initial velocity of 30 m/s at 45° above the horizontal. Calculate:

• a) The horizontal and vertical components of the initial velocity
• b) The time of flight
• c) The maximum height reached
• d) The horizontal range

Question 2: Different Angle

A soccer ball is kicked at 20 m/s at 30° above the ground. How long does it stay in the air? What's the maximum height?

Question 3: Finding the Angle

A golf ball travels a horizontal distance of 100 m when hit at 25 m/s. What was the launch angle? (Assume level ground)

Question 4: Target Practice

A cannonball is fired from a cliff that's 50 m high, with initial speed 40 m/s at 30° above horizontal. How far from the cliff base does it land?

Solutions

Solution 1: 30 m/s at 45°

a) Velocity components

v₀ₓ = 30 × cos(45°) = 30 × 0.707 = 21.2 m/s

v₀ᵧ = 30 × sin(45°) = 30 × 0.707 = 21.2 m/s

b) Time of flight

The ball goes up and comes back down. Time up = time down.

T = 2v₀ᵧ/g = 2(21.2)/9.8 = 4.33 seconds

c) Maximum height

H = v₀ᵧ²/(2g) = (21.2)²/(2 × 9.8) = 449/19.6 = 22.9 m

d) Horizontal range

R = v₀² sin(2θ)/g = (30)² × sin(90°)/9.8 = 900/9.8 = 91.8 m

Or use: R = v₀ₓ × T = 21.2 × 4.33 = 91.8 m

Solution 2: Soccer Ball at 30°

Given: v₀ = 20 m/s, θ = 30°

v₀ᵧ = 20 × sin(30°) = 20 × 0.5 = 10 m/s

Time in air:

T = 2(10)/9.8 = 2.04 seconds

Maximum height:

H = (10)²/(2 × 9.8) = 100/19.6 = 5.1 m

Solution 3: Finding the Angle

Given: R = 100 m, v₀ = 25 m/s

Use: R = v₀² sin(2θ)/g

100 = (25)² × sin(2θ)/9.8

100 = 625 × sin(2θ)/9.8

sin(2θ) = 100 × 9.8 / 625 = 980/625 = 1.568

Wait — sin(2θ) can't exceed 1. That means 100 m range isn't possible with 25 m/s on level ground. The maximum range with 25 m/s is:

Rmax = v₀²/g = 625/9.8 = 63.8 m

So either the ball was hit harder, or there's wind help, or the ball was launched from an elevated position.

Solution 4: Cannonball from a Cliff

This one's trickier. The ball starts 50 m above ground, so it takes extra time to fall.

Step 1: Find components

v₀ₓ = 40 × cos(30°) = 40 × 0.866 = 34.6 m/s

v₀ᵧ = 40 × sin(30°) = 40 × 0.5 = 20 m/s

Step 2: Set up the vertical motion equation

y = v₀ᵧ × t - ½gt²

We need y = -50 m (50 m below starting point)

-50 = 20t - ½(9.8)t²

-50 = 20t - 4.9t²

4.9t² - 20t - 50 = 0

Step 3: Solve the quadratic

t = [20 ± √(400 + 980)] / (2 × 4.9)

t = [20 ± √1480] / 9.8

t = [20 ± 38.5] / 9.8

Take the positive root: t = 58.5 / 9.8 = 5.97 seconds

Step 4: Find horizontal distance

x = v₀ₓ × t = 34.6 × 5.97 = 206.6 m

Getting Started: How to Solve Any Projectile Problem

Follow these steps in order. Skipping steps is where people lose marks.

Step 1: Draw a Diagram

Sketch the trajectory. Mark the launch point, peak, landing point, and any given heights or distances.

Step 2: List What You Know

Write down v₀, θ, g, and any distances or times given. Convert units if needed.

Step 3: Break Into Components

Find v₀ₓ and v₀ᵧ using sine and cosine. Write them down.

Step 4: Solve Horizontal and Vertical Separately

Horizontal: use x = v₀ₓ × t

Vertical: use the kinematic equations with a = -g

Step 5: Connect the Components

The time is the same for both directions. Use this to link horizontal and vertical motion.

Step 6: Plug and Solve

Substitute numbers. Solve for the unknown. Check your units.

Common Mistakes That Cost You Points

• Using the total velocity in vertical calculations. You must use the vertical component only.
• Forgetting that time of flight includes both rising and falling.
• Using the wrong sign for gravity. g = +9.8 m/s² in equations, acceleration = -9.8 m/s².
• Mixing up range and height formulas. Range uses sin(2θ), height uses sin²(θ).
• Not checking if the object lands at the same height it started. If not, you need to solve a quadratic for time.

Quick Mental Math Shortcuts

• Maximum range always occurs at 45°.
• If launch and landing heights are equal, the object spends equal time going up and coming down.
• The velocity at the peak is purely horizontal — vertical velocity is zero.
• Range is the same at θ and (90° - θ). A ball thrown at 30° lands in the same spot as one thrown at 60°.

That's it. Practice the problems above until you can solve them without checking the solutions.