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## Class 11 Physics (India)

### Course: Class 11 Physics (India)>Unit 8

Lesson 6: Projectiles launched at an angle

# 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 $90\mathrm{°}$ 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 $9.8\phantom{\rule{0.167em}{0ex}}\frac{\text{m}}{{\text{s}}^{2}}$ downward at all points of the trajectory, no matter how a projectile is launched.

### 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 $45\mathrm{°}$ 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 $9.8\phantom{\rule{0.167em}{0ex}}\frac{\text{m}}{{s}^{2}}$ (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.

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.

## Want to join the conversation?

• 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
• Do larger or smaller angles have anything to do with a larger or smaller initial velocity?
(1 vote)
• 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?
• 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.
(1 vote)
• How would you calculate initial velocity only given θ and a point of impact?
• 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...
• how does speed change when a ball is thrown?
(1 vote)
• How do you find how much time it takes to get to the maximum height and what that maximum height is?
(1 vote)
• 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
(1 vote)
• Does the initial velocity (hypotenuse) change with the angle that a projectile is shot at?
(1 vote)
• No, it doesn't.
(1 vote)
• A projectile is launched horizontally from the roof of a building that is 14 m tall. How long does it take the projectile to strike the ground?
(1 vote)
• dy = vot + 1/2at^2
We want to find t.

We know that dy (delta y) is -14, because if the horizontally launched projectile is at 0, then anything below it should be negative, hence -14m.

We know that there is no initial velocity in the y direction, the only force acting downwards is gravity:

-14 = 0t + 1/2at^2

-14 = 1/2at^2

a is the acceleration in the y direction, or gravity. This is -9.8 m/s.

So: -14 = 1/2(-9.8)t^2
Which is: -14 = -4.9t^2

Divide both sides by -4.9
-14/-4.9 = t^2
14/4.9 = t^2
take sqrt of both sides
t = sqrt(14/4.9)

ans = ~1.6903 seconds.
(1 vote)