Projectile angles and speed (x and Y) near impact

In summary, Villyer figured out that the magnitude of the velocity is 81.5 m/s and the angle is -55.7°.
  • #1
coneheadceo
25
0
Thought I was done but not the case...

Homework Statement


Alright so as stated a couple days ago... "projectile is shot from the edge of a cliff 125 m above the ground @ an initial speed of 65 m/s at an angle of 37° with the horizontal from the cliff."
---- As helped earlier I have calculated the time it takes for the projectile to hit the ground (10.4 seconds) and on my own the distance it will travel (540 M). But there is more and I am stuck...

" at the moment before impact with the ground find; 1)the horizontal and vertical components of it's velocity, 2)the magnitude of the velocity and 3) the angle made by the velocity vector with the horizontal."


Homework Equations



Not quite sure here

The Attempt at a Solution



I have figured out the Vxf by

Vxo= (65 m/s)(cos 37°)= 51.9 m/s which I equate to the velocity in the x direction for the whole flight so it can apply at impact as well as at launch.

But since the Vyf is in the downward direction I am lost as how to proceed from here to finish the rest of the problem.
 
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  • #2
To find Vfy you need to use one of the kinematic equations for constant acceleration and apply it on the vertical component of velocity.
 
  • #3
So would I go something like this?

Vyf = Vyo + at

Vyf = (65 m/s)(sin 37) + (-9.8 m/s^2)(10.4s)

= 39.1 m/s - 101.92 m/s

= -62.82 m/s

answer in the book is -63.1 m/s so I am comfortable with that.

So the overall magnitude of the velocity I would go...

√[(51.9)^2 + (-62.8)^2
= 81.5 m/s

for the overall angle on that beast would I then use...

sin θ = 51.9/ -62.8
= -55.7°

Book says 50.6 below, which is understood because it is negative but I am off with something here.. what am I missing?
 
  • #4
coneheadceo said:
sin θ = 51.9/ -62.8

Take a second look at this.
 
  • #5
Flip it?

so go...

tan θ = -62.8/51.9

= -50.4°

Think that will jive... thank you for the help Villyer!
 
  • #6
You're welcome! And it wasn't just flipping it, you also wrote sin the first time:b
 

FAQ: Projectile angles and speed (x and Y) near impact

1. What is the relationship between projectile angle and speed near impact?

The relationship between projectile angle and speed near impact is known as the trajectory of the projectile. This trajectory is affected by both the initial angle at which the projectile is launched and its initial speed. As the projectile nears its impact point, its speed and angle will both change, ultimately determining the exact point of impact.

2. How does the angle of impact affect the distance traveled by the projectile?

The angle of impact can significantly affect the distance traveled by the projectile. A shallower angle of impact will result in a longer distance traveled, as the projectile will have a more horizontal trajectory. On the other hand, a steeper angle of impact will result in a shorter distance traveled, as the projectile will have a more vertical trajectory.

3. How does the speed of the projectile near impact affect its impact force?

The speed of the projectile near impact has a direct correlation with its impact force. The higher the speed, the greater the impact force will be. This is because a higher speed means the projectile has more kinetic energy, which is converted into impact force upon impact.

4. Can the angle and speed of a projectile be manipulated to achieve a desired impact point?

Yes, the angle and speed of a projectile can be manipulated to achieve a desired impact point. By adjusting the initial angle and speed, the trajectory of the projectile can be altered to reach a specific point of impact. This is often seen in sports such as archery or javelin throwing.

5. How does air resistance affect the angle and speed of a projectile near impact?

Air resistance can have a significant impact on the angle and speed of a projectile near impact. As the projectile moves through the air, it experiences drag, which can slow it down and alter its trajectory. This means that the angle and speed at which the projectile was launched may not be the same as when it reaches its impact point, as they are affected by air resistance.

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