Finding initial velocity given angle and horizontal displacement

In summary, the homework statement states that the distance between the object and the observer is 0.8 meters, and the angle is 55 degrees. Assuming that the object leaves and falls back on the same horizontal plane, the gravitational force is 9.81 meters/second squared.
  • #1
Suprin
24
0

Homework Statement


distance = 0.8meters
angle = 55°
gravity = [itex]9.81m/s^2[/itex]
[itex]\V_0[/itex] = ?, assuming that it leaves and falls back on the same horizontal plane.
The attempt at a solution

So we know for a fact that the initial velocity is supposed to be [itex]2 m/s[/itex], according to the book at least.

I tried using the formula [itex] R = \frac{V_0^2 sin2\theta}{g} [/itex] but that gives me a completely different number.

I've discarded the use of component formulas since I'd require initial velocity for those. I'd like to say that I believe that the initial velocity on X is different to the initial velocity on Y since it's not 45°, but I get the feeling that doesn't matter. I'm stuck and frustrated on this relatively simple problem and can't figure it out for the life of me.

I'm also trying to get the hang of this Latex thing :p
 
Last edited:
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  • #2
The formula you have for range is wrong, which is obvious if you check the units. Use the correct formula and you should be fine.
 
  • #3
If you meant the [itex]V_0^2[/itex], I fixed it. That was a typo. Still not getting the right answer.
 
  • #4
Use the equations for the x and y positions for projectile motion and use the final positions for x and y.
 
  • #5
Suprin said:
If you meant the [itex]V_0^2[/itex], I fixed it. That was a typo. Still not getting the right answer.

Show exactly what you do.
 
  • #6
voko said:
Show exactly what you do.

I have no idea what I did wrong before, but as soon as I was about to finish copying the last step to reply to you, I got the answer.

It was one of those "watch which buttons you press on the calculator" things. Sorry :/

Rounding at the very end, I do get 2.9m/s.
 
  • #7
No need to be sorry, you solved the problem, and I am glad I was able to help, even if only by having you do that again :)
 

Related to Finding initial velocity given angle and horizontal displacement

1. What is the formula for finding initial velocity given angle and horizontal displacement?

The formula for finding initial velocity given angle and horizontal displacement is:
V0 = (dX * g) / (sin(2 * θ))
Where V0 is the initial velocity, dX is the horizontal displacement, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of launch.

2. Can this formula be used for any angle of launch?

Yes, this formula can be used for any angle of launch as long as the angle is measured in radians. If the angle is given in degrees, it must be converted to radians first by multiplying by π/180.

3. What are the units for initial velocity when using this formula?

The units for initial velocity when using this formula are meters per second (m/s). This is because the units for horizontal displacement (m), acceleration due to gravity (m/s^2), and angle (radians) all cancel out, leaving meters per second as the final unit.

4. Is this formula only applicable for objects launched horizontally?

No, this formula can be used for objects launched at any angle, as long as the horizontal displacement and angle are known. It takes into account the horizontal and vertical components of the initial velocity.

5. What are the limitations of using this formula to find initial velocity?

This formula assumes that there is no air resistance and that the acceleration due to gravity is constant. It also assumes that the launch and landing points are at the same height. In real-life situations, these assumptions may not hold true and can affect the accuracy of the calculated initial velocity.

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