If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Projectiles launched at an angle review

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

Key terms

TermMeaning
Launch angleThe angle of a projectile’s initial velocity when measured from the horizontal direction. These angles are typically 90, degree 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, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction downward at all points of the trajectory, no matter how a projectile is launched.
In the center is a horizontal black line. An arrow arcs rightward as a light blue parabola from one end of the parabola to the other. There are 5 small dark blue dots evenly spaced along the arc. Each dot has an identical purple arrow pointing downwards. The arrows are labeled "a = -g".
Figure 1. Acceleration a is constant, downwards, and equal to g

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.
In the center is a horizontal black line. An arrow arcs rightward as a light blue parabola from one end of the parabola to the other. There are 5 small dark blue dots evenly spaced along the arc. Each dot has an identical purple arrow pointing rightwards. The arrows are labeled "v_x".
Figure 2. Horizontal velocity is constant

Vertical velocity changes direction and magnitude during trajectory

Before the object reaches the maximum height, the vertical speed v, start subscript, y, end subscript 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
In the center is a horizontal black line. An arrow arcs rightward as a light blue parabola from one end of the parabola to the other. There are 5 small dark blue dots evenly spaced along the arc. Each dot has purple vertical arrow. On the left side of the arc, the purple arrows point up and reduce to zero length at the top. On the right side of the arc, the purple arrows point down and increase towards the right.
Figure 3. A projectile’s constant downward acceleration a changes the vertical velocity v, start subscript, y, end subscript throughout the trajectory

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.
There are three right triangles. Each triangle's hypotenuse is a thick purple arrow, while its two legs are dashed purple lines. At the lower left vertex of each triangle is a blue dot. The hypotenuse of the first triangle is labeled "v sub o", its horizontal leg is labeled "v sub o x" and its vertical leg is labeled "v sub o y".
All three triangles' hypotenuses are the same length. The first triangle has its left vertex at a 45 degree angle, the second triangle at a 30 degree angle, and the third triangle at a 60 degree angle. As the angle changes, the lengths of the triangles' horizontal and vertical legs change too.
Figure 4. Changing the launch angles changes the initial velocity components

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.
Along the images bottom is a horizontal black line. At the left of this black line is a cannon. From the barrel of the cannon, three dashed arrows shaped as parabolas originate and arc to the right of the black line. One purple parabola starts at 60 degree angle and is tall and terminates at 3/4 the way across the black line. Another purple parabola starts at 30 degree angle and is short and terminates at 3/4 the way across the black line also. A red parabola starts at 45 degree angle and is tall and terminates at the end of the black line.
Figure 5: Trajectories of launch angles with the same initial speed

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, degree 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, point, 8, start fraction, start text, m, end text, divided by, s, squared, end fraction (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.

Want to join the conversation?

  • blobby green style avatar for user preny08
    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?
    (14 votes)
    Default Khan Academy avatar avatar for user
    • mr pants orange style avatar for user Natalie Semerky
      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.)
      (16 votes)
  • aqualine ultimate style avatar for user Katie
    what would be a good equation for solving range(x) if I only have angle that the projectile was launched?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • hopper cool style avatar for user obiwan kenobi
      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!
      (4 votes)
  • blobby green style avatar for user Keerthana
    Do larger or smaller angles have anything to do with a larger or smaller initial velocity?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Sophie Dowling
    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?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • piceratops seed style avatar for user Kevin Liang
    For the question of comparing the horizontal distance traveled of different initial angles of projection. I tried to drive a formula, ending up having the horizontal distance traveled = v^2sin(2θ)/g. Hence the optimal angle of projection for the greatest horizontal distance is 45° because sin(90) = 1, and any other angle will result in a value smaller than 1.

    1. Knowing that the horizontal velocity = vcos(θ), so we can get the horizontal distance(s) = horizontal velocity x time, s = vcos(θ)t.

    2. So the issue is to find time(t), the time is affected by the vertical component of velocity and the acceleration due to gravity(g). Knowing that the time it takes for the projectile to reach the maximum height from its initial height is the same as the time it takes to fall from the maximum height back to its initial height. At maximum height, the vertical velocity(vsin(θ)) is reduced to zero, so the equation should give vsin(θ) - gt = 0. Rearranging the equation for finding t, vsin(θ)/g = t, this is the time it takes to reach its maximum height, so we multiply by 2 to get the total time for it to reach the maximum height and return back to the initial height. 2vsin(θ)/g = t(total).

    3. From the equation s = vcos(θ)t, and t = 2vsin(θ)/g. Subsititing the equation, getting s = 2v^2sin(θ)cos(θ)/g. Lets remind us about the trigonometry identity sin(2θ) = 2cos(θ)sin(θ). Hence the equation can be simplified to s = v^2sin(2θ)/g.

    However, this only works for the scenario that the direction of gravity is always one direction that is vertically downwards. A more realistic scenario is having the direction of gravity towards a center, which is definitely much harder to derive such an equation, and also you will have to redefine the distance traveled as Δθr, assuming that Earth is a perfect sphere with radius(r).
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Julieta Jara, s1531670
    is horizontal distance the same as horizontal range? I have a problem that wants me to find horizontal range but I am only given initial velocity (10 m/s) and height (8 m)
    (2 votes)
    Default Khan Academy avatar avatar for user
    • hopper happy style avatar for user Averell Chen
      The horizontal distance travelled by a projectile is called its range. A projectile launched on level ground with an initial speed v0 at an angle θ above the horizontal will have the same range as a projectile launched with an initial speed v0 at 90° − θ and maximum range when θ = 45°.
      (1 vote)
  • blobby green style avatar for user Alicia Sanchez
    Can velocity have a degree component? Or is the true bearing notation only for displacement?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • hopper happy style avatar for user Averell Chen
    "The only acceleration of a projectile is the downwards acceleration due to gravity". Is there no acceleration as the projectile is being launched forward or does that count as velocity?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • aqualine ultimate style avatar for user Welton
    in all these cases, is it only if the object was in a väcuum?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • duskpin sapling style avatar for user Sage Luong
      Yes, this is assuming that the object is in a vacuum. Otherwise, the force of air friction would impede the projectile's horizontal motion, thus causing it to decelerate. In projectile motion, the horizontal acceleration is 0.
      (2 votes)
  • piceratops seedling style avatar for user djohnson.sp
    How would you calculate initial velocity only given θ and a point of impact?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Obada Amer
      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...
      (0 votes)