Why is 45 degrees the angle at which horizontal displacement is greatest? Wouldn't a horizontal launch have the greatest displacement since the object will be traveling in a straight path?

45 degrees optimizes the horizontal distance hypothetically. On Earth, this value is closer to 42 degrees. To echo what Prince Grey said, a horizontally launched projectile would be immediately pulled to the ground by gravity. (This assumes you are on a planet with gravity like Earth.)

what would be a good equation for solving range(x) if I only have angle that the projectile was launched?

You would have to create a system of equations: v0 sin(theta) - 9.8t = 0 (In this equation t is the time to max height). 2(-9.8)t^2 + 2*v0*t sin (theta) = 0 (In this equation t is also the time to max height) Note: These equations come straight from the kinematic equations and 2*t is only the total time in the air if the projectile lands at the same level from which it is launched. Hope this helps!

How do you find the maximum height?

to find the H max, or the highest height u^2 sin^2(theta) _______________ 2g

Let's say you launch a ball of a 100m tall cliff and an angle. How do I know when to plug the displacement of 100m in kinematics equations as negative or positive?

Usually, when dealing with displacement problems, a negative value is used for downward displacement. With your 100m cliff example, this would be the case. Negative displacement would also occur when moving an object to the left in a 2-dimensional space. Positive, of course, would be the opposite. I hope that answered your question, but if it didn't... Displacement is the total distance from the starting position. Even if the arc of the projectile began by moving upward, if it ends below its original position, it has a negative displacement.

Main content

Course: High school physics > Unit 3

Lesson 5: Projectiles launched at an angle

Projectiles launched at an angle review

Google Classroom

Learn about projectile motion vectors and how the launch angle impacts the trajectory.

Key terms

Term	Meaning
Launch angle	The angle of a projectile’s initial velocity when measured from the horizontal direction. These angles are typically $90 °$ ‍ or less.

Vectors of projectiles launched at an angle

Constant vertical acceleration

The only acceleration of a projectile is the downwards acceleration due to gravity (see Figure 1 below). Vertical acceleration is always equal to

9.8 \frac{m}{s^{2}}

downward at all points of the trajectory, no matter how a projectile is launched.

All objects moving through the air experience some amount of air resistance. However, dealing with air resistance can be tricky, so we'll typically assume that the force of air resistance is negligible (i.e., small enough to be ignored). This is not always a good approximation, but it works really well for objects whose weights are large compared to the force of air resistance.

Keep in mind that the force of air resistance on an object gets larger and larger as the speed of the object increases. So we'll typically consider situations where the speed of an object is small enough that air resistance is still negligible.

No horizontal acceleration

Nothing accelerates a projectile horizontally, so horizontal acceleration is always zero.

Horizontal velocity is constant

The projectile’s horizontal speed is constant throughout the entire trajectory (see figure 2 below) because gravity only acts downwards in the vertical direction.

Vertical velocity changes direction and magnitude during trajectory

Before the object reaches the maximum height, the vertical speed

v_{y}

of a projectile decreases, because acceleration is in the opposite direction. The direction of the velocity is initially upward, since the object’s height is increasing (see Figure 3 below).

Vertical velocity becomes zero at the projectile’s maximum height. The vertical speed increases after the maximum height because acceleration is in the same direction (see figure 3 below). The direction of vertical velocity is downward as the object’s height decreases

Analyzing angled launch trajectories

Components of initial velocity

To see how to break down the total velocity vector into the horizontal and vertical components using trigonometry, see the article on analyzing vectors.

Launch angle trajectory comparisons

The diagram below shows trajectories for different launch angles that have the same initial speed. The launch angle determines the maximum height, time in the air, and maximum horizontal distance of the projectile.

Higher launch angles have higher maximum height

The maximum height is determined by the initial vertical velocity. Since steeper launch angles have a larger vertical velocity component, increasing the launch angle increases the maximum height. (see figure 5 above).

Higher launch angles have greater times in the air

The time in air is determined by the initial vertical velocity. Since steeper launch angles have a larger vertical velocity component, increasing the launch angle increases the time in air. For deeper explanations of the relationship between projectile time in air and initial vertical velocity, see Sal’s video on the optimal angle for a projectile.

Projectile maximum horizontal distance depends on horizontal velocity and time in air

Launch angles closer to

45 °

give longer maximum horizontal distance (range) if initial speed is the same (see figure 5 above). These launches have a better balance of the initial velocity components that optimize the horizontal velocity and time in air (see figure 4).

Common misconceptions

People mix up horizontal vs. vertical components of acceleration and velocity. The acceleration is a constant downwards $9.8 \frac{m}{s^{2}}$ ‍ (see figure 1) because gravity is the only source of acceleration. This acceleration only changes the vertical velocity, so the horizontal velocity is constant.
People can’t remember what is zero at the maximum height. Vertical velocity is zero at this point, but there is still horizontal velocity and acceleration is still down.

Learn more

For deeper explanations, see our video about finding the optimal launch angle for a projectile and our video on finding the horizontal displacement for a projectile launched at an angle.

To check your understanding and work toward mastering these concepts, check out these exercises: analyzing vectors for projectile trajectories and vectors and comparing multiple trajectories.

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preny08
Posted 6 years ago. Direct link to preny08's post “Why is 45 degrees the ang...”
Why is 45 degrees the angle at which horizontal displacement is greatest? Wouldn't a horizontal launch have the greatest displacement since the object will be traveling in a straight path?
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(15 votes)
Answer
- Natalie Semerky
  Posted 6 years ago. Direct link to Natalie Semerky's post “45 degrees optimizes the ...”
  45 degrees optimizes the horizontal distance hypothetically. On Earth, this value is closer to 42 degrees. To echo what Prince Grey said, a horizontally launched projectile would be immediately pulled to the ground by gravity. (This assumes you are on a planet with gravity like Earth.)
  Button navigates to signup page
  (19 votes)
Katie
Posted 5 years ago. Direct link to Katie's post “what would be a good equa...”
what would be a good equation for solving range(x) if I only have angle that the projectile was launched?
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(2 votes)
Answer
- obiwan kenobi
  Posted 5 years ago. Direct link to obiwan kenobi's post “You would have to create ...”
  You would have to create a system of equations:
  
  v0 sin(theta) - 9.8t = 0 (In this equation t is the time to max height).
  2(-9.8)t^2 + 2*v0*t sin (theta) = 0 (In this equation t is also the time to max height)
  
  Note: These equations come straight from the kinematic equations and 2*t is only the total time in the air if the projectile lands at the same level from which it is launched. Hope this helps!
  Comment on obiwan kenobi's post “You would have to create ...”
  (4 votes)
skaplan2023
Posted 2 years ago. Direct link to skaplan2023's post “How do you find the maxim...”
How do you find the maximum height?
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(2 votes)
Answer
- Sukhada
  Posted 10 months ago. Direct link to Sukhada's post “to find the H max, or the...”
  to find the H max, or the highest height
  u^2 sin^2(theta)
  _____________
  2g
  Button navigates to signup page
  (3 votes)
anna.mt.ly
Posted 6 months ago. Direct link to anna.mt.ly's post “Let's say you launch a ba...”
Let's say you launch a ball of a 100m tall cliff and an angle. How do I know when to plug the displacement of 100m in kinematics equations as negative or positive?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- FlipWing
  Posted 6 months ago. Direct link to FlipWing's post “Usually, when dealing wit...”
  Usually, when dealing with displacement problems, a negative value is used for downward displacement. With your 100m cliff example, this would be the case. Negative displacement would also occur when moving an object to the left in a 2-dimensional space. Positive, of course, would be the opposite.
  
  I hope that answered your question, but if it didn't...
  
  Displacement is the total distance from the starting position. Even if the arc of the projectile began by moving upward, if it ends below its original position, it has a negative displacement.
  Button navigates to signup page
  (4 votes)
Keerthana
Posted 5 years ago. Direct link to Keerthana's post “Do larger or smaller angl...”
Do larger or smaller angles have anything to do with a larger or smaller initial velocity?
Button navigates to signup pageComment on Keerthana's post “Do larger or smaller angl...”
(1 vote)
Answer
Sophie Dowling
Posted 4 years ago. Direct link to Sophie Dowling's post “if you changed the mass o...”
if you changed the mass of the projectile (keeping the same size ball or something) how do you prove that the range will change with it?
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(2 votes)
Answer
- obiwan kenobi
  Posted 4 years ago. Direct link to obiwan kenobi's post “The thing is that if you ...”
  The thing is that if you have two projectiles that are launched with the same velocity at the same the angle, the horizontal range that the projectiles cover is actually independent of the mass. A 1 kg ball will travel the same distance as a 1000 kg ball.
  Comment on obiwan kenobi's post “The thing is that if you ...”
  (1 vote)
djohnson.sp
Posted 5 years ago. Direct link to djohnson.sp's post “How would you calculate i...”
How would you calculate initial velocity only given θ and a point of impact?
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(2 votes)
Answer
- Obada Amer
  Posted 4 years ago. Direct link to Obada Amer's post “Yes!! Use the formula v^2...”
  Yes!! Use the formula v^2=2*acceleration(which is gravity)*distance.
  
  Lets take an example
  
  Lets say the angle is 45 and the distance is 20m. Use the formula above and you'll get 19.8m/s as the speed. It's a nice question though...
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  (0 votes)
avatesoriero
Posted 7 months ago. Direct link to avatesoriero's post “how does speed change whe...”
how does speed change when a ball is thrown?
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(1 vote)
Answer
Arsh
Posted 4 years ago. Direct link to Arsh's post “How do you find how much ...”
How do you find how much time it takes to get to the maximum height and what that maximum height is?
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(1 vote)
Answer
- Sukhada
  Posted 10 months ago. Direct link to Sukhada's post “There's a special formula...”
  There's a special formula for both of these!
  To find the time taken by the projectile, we calculate-
  [2usin(theta)]
  _____________
  g
  where g is 9.8m/s2,
  u is the initial velocity, and theta is the angle of the projectile.
  
  to find the H max, or the highest height
  u^2 sin^2(theta)
  _______________
  2g
  Button navigates to signup page
  (1 vote)
Wu, Ryan
Posted 2 years ago. Direct link to Wu, Ryan's post “Does the initial velocity...”
Does the initial velocity (hypotenuse) change with the angle that a projectile is shot at?
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(1 vote)
Answer
- rylan.wetsell
  Posted a year ago. Direct link to rylan.wetsell's post “No, it doesn't.”
  No, it doesn't.
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  (1 vote)