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### Course: High school physics>Unit 3

Lesson 3: Graphs of projectile motion

# Projectile motion graphs

Visualizing position, velocity and acceleration in two-dimensions for projectile motion.

## Want to join the conversation?

• Why is the second and third Vx are higher than the first one?
• All thanks to the angle and trigonometry magic.
After looking at the angle between actual velocity vector and the horizontal component of this velocity vector, we can state that:
1) in the second (blue) scenario this angle is zero;
2) in the third (yellow) scenario this angle is smaller than in the first scenario.

If above described makes sense, now we turn to finding velocity component. We do this by using cosine function: cosine = horizontal component / velocity vector. After manipulating it, we get something that explains everything!
horizontal component = cosine * velocity vector

Now we get back to our observations about the magnitudes of the angles. If we work with angles which are less than 90 degrees, then we can infer from unit circle that the smaller the angle, the higher the value of its cosine.
So from our derived equation (horizontal component = cosine * velocity vector) we get that the higher the value of cosine, the higher the value of horizontal component (important note: this works provided that velocity vector has the same magnitude. In this case, this assumption (identical magnitude of velocity vector) is correct and is the one that Sal makes, too).

Now let's get back to our observations:
1) in blue scenario, the angle is zero; hence, cosine=1. This means that the horizontal component is equal to actual velocity vector. It actually can be seen - velocity vector is completely horizontal.
2) in yellow scenario, the angle is smaller than the angle in the first (red) scenario. This means that cos(angle, red scenario) < cos(angle, yellow scenario)! Hence, the horizontal component in the third (yellow) scenario is higher in value than the horizontal component in the first (red) scenario.
• At in the top diagram, shouldn't the ball have a little positive acceleration as if was in state of rest and then we provided it with some velocity?
• maybe have a positive acceleration just before into air, once the ball out of your hand, there will be no force continue exerting on it, except gravitational force (assume air resistance is negligible), so in the whole journey only gravity affect acceleration.
• Why did Sal say that v(x) for the 3rd scenario (throwing downward -orange) is more similar to the 2nd scenario (throwing horizontally - blue) than the 1st (throwing upward - "salmon")? I would have thought the 1st and 3rd scenarios would have more in common as they both have v(y)>0.
• how the velocity along x direction be similar in both 2nd and 3rd condition?
(that is in blue and yellow)
• Because there are no forces in that direction.
• At , how is the blue graph's x initial velocity a little bit more than the red graph's x initial velocity?
• In this case/graph, we are talking about velocity along x- axis(Horizontal direction)
So, initial velocity= u cosӨ
For blue ball and for red ball Ө(angle with which the ball is projected) is different(it is 0 degrees for blue, and some angle more than 0 for red)

Now, assuming that the two balls are projected with same |initial velocity| (say u), then the initial velocity will only depend on cosӨ in initial velocity = u cosӨ, because u is same for both.

Now, let's see whose initial velocity will be more -
It'll be the one for which cos Ө will be more
For blue, cosӨ= cos0 = 1
For red, cosӨ= cos (some angle>0)= some value,say x<1
[because we know that as Ө increases, cosӨ decreases
cos(0)=1
therefore, cos(Ө>0)=x<1]

Now, we have,
Initial velocity of blue ball = u cosӨ = u*(1)= u
Initial velocity of red ball = u cosӨ = u*(x<1)= some value, say y<u
Therefore, initial velocity of blue ball> initial velocity of red ball
Hence, Sal plots blue graph's x initial velocity(initial velocity along x-axis or horizontal axis) a little bit more than the red graph's x initial velocity(initial velocity along x-axis or horizontal axis).
(1 vote)
• I cannot understand why the acceleration in x direction is zero in all three cases? If we throw something in the air, we put force on it. So the acceleration in both directions should be made.
• Gravity acts in a downward direction. In actuality there is wind resistance acting horizontally against the object, but for the scope of introducing the concept, it is important to recognize that in the absence of wind resistance, there is no force acting horizontally on the object
• Hi there, at why does Sal draw the graph of the orange line at the same place as the blue line? I thought the orange line should be drawn at the same level as the red line. Thanks!
• why is the acceleration of the y-value negative? Why is the acceleration of the x-value 0.