How to Find the Range and Maximum Height of a Projectile Launched at an Angle

The Problem

Find the range and maximum height:

A ball is launched from ground level with an initial velocity of 30 m/s at an angle of 45° above the horizontal. Find (a) the maximum height reached by the ball, and (b) the horizontal range of the projectile.

Assume air resistance is negligible and use g = 9.8 m/s².

This is one of those classic projectile motion problems you'll see in every introductory physics course. You've got an object launched at an angle, and you need to figure out how high it goes and how far it travels.

The key here is recognizing that projectile motion is just two independent motions happening at the same time: constant velocity horizontally and constant acceleration vertically. If you're stuck on a similar problem or need to see the calculation worked out in detail, you can always generate a custom video solution on Torial.

Understanding Projectile Motion

Projectile motion diagram showing velocity components and trajectory

Projectile motion is the motion of an object launched into the air, subject only to gravity. The path it follows is a parabola.

Key principles:

Horizontal motion: Constant velocity (no acceleration in x-direction)
Vertical motion: Constant acceleration downward (g = 9.8 m/s²)
Independence: These two motions are completely independent
Time: The same time applies to both horizontal and vertical motion

💡 Key Insight: The horizontal and vertical motions share the same time variable. When the projectile reaches its maximum height, the vertical velocity is zero, but the horizontal velocity remains constant throughout the flight.

This separation makes the problem solvable. You can solve for time using vertical motion, then use that same time to find horizontal distance.

Think about it: if you throw a ball, it goes up and comes down while simultaneously moving forward. The up-and-down part is affected by gravity, but the forward part isn't. That's why we can treat them separately. Need more help visualizing projectile motion? Check out other physics videos in our library.

Breaking Velocity into Components

The first step is always to break the initial velocity into its horizontal and vertical components using trigonometry.

Given information:

Initial velocity (v₀):

30 m/s

Launch angle (θ):

45°

Step 1: Find the horizontal component (v₀ₓ)

Using cosine:

v₀ₓ = v₀ cos θ = 30 cos(45°) = 30 × (√2/2) = 30 × 0.7071 ≈ 21.21 m/s

This is the constant horizontal velocity throughout the flight.

Step 2: Find the vertical component (v₀ᵧ)

Using sine:

v₀ᵧ = v₀ sin θ = 30 sin(45°) = 30 × (√2/2) = 30 × 0.7071 ≈ 21.21 m/s

This is the initial vertical velocity. It will decrease due to gravity.

Note: For a 45° angle, the horizontal and vertical components are equal. This is actually the optimal angle for maximum range (when launched from ground level).

Now that we have the components, we can solve for maximum height using vertical motion, then use the time of flight to find the range.

Finding Maximum Height

Maximum height calculation for projectile motion

At the maximum height, the vertical velocity becomes zero. We can use the kinematic equation for vertical motion.

Kinematic equation:

v² = v₀² + 2aΔy

Where v is final velocity, v₀ is initial velocity, a is acceleration, and Δy is displacement.

At maximum height:

vᵧ = 0 (vertical velocity is zero at the peak)
v₀ᵧ = 21.21 m/s (initial vertical velocity)
a = -g = -9.8 m/s² (acceleration due to gravity, negative because it's downward)
Δy = h (the maximum height we're solving for)

0² = (21.21)² + 2(-9.8)h

0 = 450 - 19.6h

19.6h = 450

h = 450 / 19.6 ≈ 22.96 m

Answer (a): Maximum Height

h = 23.0 m

(rounded to 3 significant figures)

The ball reaches a maximum height of approximately 23 meters. Notice that we used the vertical component of velocity and vertical acceleration. The horizontal motion doesn't affect the maximum height at all.

Calculating Time of Flight

To find the range, we need to know how long the projectile is in the air. We can find this using vertical motion.

Method 1: Using time to reach maximum height

At maximum height, vᵧ = 0. Using v = v₀ + at:

0 = 21.21 + (-9.8)t

9.8t = 21.21

t = 21.21 / 9.8 ≈ 2.16 s

This is the time to reach maximum height. Since the path is symmetric, the total time of flight is twice this:

T = 2 × 2.16 = 4.32 s

Method 2: Using the full trajectory

Using y = v₀ᵧt + ½at², with y = 0 (returns to ground level):

0 = (21.21)t + ½(-9.8)t²

0 = 21.21t - 4.9t²

0 = t(21.21 - 4.9t)

t = 0 or t = 21.21 / 4.9 ≈ 4.32 s

The t = 0 solution is the launch time. The t = 4.32 s is when it returns to the ground.

Both methods give the same answer: Total time of flight T = 4.32 s. This is the time we'll use to calculate the horizontal range.

Either method works. The first method uses the symmetry of the trajectory, while the second directly solves for when the projectile returns to ground level. Both give you the same time.

Finding the Horizontal Range

Now that we know the time of flight, finding the range is straightforward. Horizontal motion has constant velocity.

Horizontal motion equation:

x = v₀ₓt

Since there's no horizontal acceleration, we just multiply horizontal velocity by time.

Calculating the range:

R = v₀ₓ × T

R = 21.21 × 4.32

R ≈ 91.6 m

The horizontal range is approximately 91.6 meters.

Answer (b): Horizontal Range

R = 91.6 m

(rounded to 3 significant figures)

💡 Quick check: For a 45° launch angle from ground level, the range formula is R = v₀²/g. Let's verify:

R = (30)² / 9.8 = 900 / 9.8 ≈ 91.8 m

This matches our calculation! The slight difference is due to rounding in intermediate steps.

The projectile travels approximately 91.6 meters horizontally before hitting the ground. Notice how we used the horizontal velocity (which stays constant) and the total time of flight (which we found from vertical motion). Want to see more worked examples? Browse through hundreds of physics solutions on Torial.

Common Mistakes to Avoid

Here are the mistakes that cost students the most points. Learn them now so you don't make them on test day. If you want personalized help avoiding these errors, create a custom study video for your specific problem.

❌ Mistake #1: Using the full velocity instead of components

Plugging v₀ = 30 m/s directly into vertical motion equations. The 30 m/s is at an angle, not straight up.

Fix: Always break velocity into components first. Use v₀ᵧ = v₀ sin θ for vertical motion and v₀ₓ = v₀ cos θ for horizontal motion.

❌ Mistake #2: Forgetting that acceleration is negative

Using a = +9.8 m/s² instead of a = -9.8 m/s² in vertical motion equations. Gravity points downward.

Fix: Define your coordinate system. If up is positive, then g = -9.8 m/s². Always be consistent with your sign convention.

❌ Mistake #3: Using the wrong time for range calculation

Using the time to reach maximum height (2.16 s) instead of total time of flight (4.32 s) when calculating range.

Fix: The range uses the total time the projectile is in the air, not just the time to the peak. Remember: the projectile has to come back down.

❌ Mistake #4: Mixing up sin and cos

Using v₀ₓ = v₀ sin θ and v₀ᵧ = v₀ cos θ. This swaps horizontal and vertical components.

Fix: Remember: horizontal uses cosine (adjacent side), vertical uses sine (opposite side). Draw a triangle if you're unsure.

❌ Mistake #5: Assuming horizontal acceleration

Using x = v₀ₓt + ½at² with a = -9.8 for horizontal motion. There's no horizontal acceleration in projectile motion.

Fix: Horizontal motion has constant velocity. Use x = v₀ₓt. The only acceleration is vertical (gravity).

❌ Mistake #6: Not using the correct kinematic equation

Trying to use v = v₀ + at to find maximum height directly. You need v² = v₀² + 2aΔy when you know the final velocity is zero.

Fix: Choose the right equation based on what you know. For maximum height, you know v = 0, so v² = v₀² + 2aΔy is perfect.

Practice Problems with Video Solutions

Best way to get good at this? Practice. Try these similar problems and check your work with video solutions. You can also generate instant video explanations for any projectile motion problem you're working on.

Practice Problem 1: Different Angle

A projectile is launched with an initial velocity of 25 m/s at 30° above the horizontal. Find the maximum height and range.

Hint: Follow the same steps. Break into components, find maximum height, then time of flight, then range.

Get instant video solution on Torial →

Practice Problem 2: Launched from Height

A ball is thrown from a height of 10 m with velocity 20 m/s at 60° above horizontal. Find where it lands.

Hint: The initial vertical position is not zero. Adjust your equations accordingly.

Get instant video solution on Torial →

When to Use Component Method vs. Range Formula

Should you always break into components, or can you use the range formula directly?

✓ Use Component Method When:

You need maximum height (not just range)
The launch point is not at ground level
You need intermediate times or positions
The problem asks for step-by-step work

✓ Use Range Formula When:

You only need the range
Launch is from ground level
You want a quick answer
The problem is straightforward

For this problem? Component method is best because we need both maximum height and range. When you're working with projectile motion, it helps to have a step-by-step video walkthrough that shows exactly which method to use and when.

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