Difficult Projectile problem

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In summary: You can also use the fact that sin(45)=cos(45)=0.707, but I'm sure you'll find it easier to simplify it if you cancel out Vix and Viy. In summary, the question is asking for the initial horizontal component of a golf ball's velocity, assuming no air resistance and an optimum angle of 45 degrees. The given distance of 418.78 meters is the horizontal range. By using the formula for range and eliminating the horizontal acceleration, we can calculate the initial horizontal velocity. Then, using the equation for vertical displacement and the fact that the horizontal and vertical components of the initial velocity are equal, we can solve for the initial horizontal velocity.
  • #1
fro
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The question is:
"A golf ball travels 418.78 meters. Assuming no air resistance and that the ball was shot at an optimum angle, what was the initial horizontal component of the ball's velocity?"

I am assuming 45 degrees to be the optimum angle. I am unsure if 418.78m is the length of the adjacent side (horizontal distance) or the length of the hypotenuse?

If I take 418.78m as the length of the hypotenuse, then the horizontal distance should be 296.122m (418.78 X cosine 45).

Since I do not know the time or the initial velocity, how can I solve this problem?

Thanks in advance for your help.
 
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  • #2
418.78 is the horizontal range. Now, since you know the range and the projection angle, you can caluculate the initial velocity using the formula for range.
 
  • #3
fro said:
The question is:
"A golf ball travels 418.78 meters. Assuming no air resistance and that the ball was shot at an optimum angle, what was the initial horizontal component of the ball's velocity?"

I am assuming 45 degrees to be the optimum angle. I am unsure if 418.78m is the length of the adjacent side (horizontal distance) or the length of the hypotenuse?

If I take 418.78m as the length of the hypotenuse, then the horizontal distance should be 296.122m (418.78 X cosine 45).

Since I do not know the time or the initial velocity, how can I solve this problem?

Thanks in advance for your help.

You do not have to know the time. The total distance that the ball traveled is enough. Use the equation for displacement in the x-direction, which is [tex]x(t)=v_{0x}\cos(45)\cdot t[/tex]. As stated, you know the distance, so plug it into the equation. The next equation you need to use is the equation of vertical displacement, i.e. in the y-direction, which is [tex]y(t)=v_{0y}\sin(45)\cdot t-\frac{1}{2}\cdot g \cdot t^2[/tex]. Further on, you know what y equals when the ball reaches its final distance, i.e. when it falls onto the ground. I'm sure you know how to proceed now.(Hint: eliminate the time t from the equation.)
 
  • #4
neutrino said:
418.78 is the horizontal range. Now, since you know the range and the projection angle, you can caluculate the initial velocity using the formula for range.

So,
418.78m = Vx X time ?

Vx = initial horizontal velocity

Since horizontal acceleration is 0, 1/2*a*t^2 is also 0 and can be left out.
 
  • #5
fro said:
So,
418.78m = Vx X time ?

Vx = initial horizontal velocity

Since horizontal acceleration is 0, 1/2*a*t^2 is also 0 and can be left out.
If you've not come across the formula I mentioned before, I guess it's better to use the method suggested by radou. :)
 
  • #6
@radou

418.78 = Vix * cos(45) * t
418.78 = Vix * 0.707 * t
t = 592.24/Vix.

Can I plug the above value of t in the y-displacement equation?

If so then,
0 = Viy * sin(45) * (592.24/Vix) * - [0.5 * g * (592.24/Vix)^2]
0 = Viy * 0.707 * (592.24/Vix) - (1753741.088/Vix^2)

Can I cancel out the Viy and Vix since I know that the vertical and horizontal components must be equal (because it is 45 degress)?
 
  • #7
It's not Vix cos(45) , Vix = Vi cos(45), where Vi is the initial velocity. But you're right, Vix = Viy, in this case.
 
  • #8
fro said:
...Can I cancel out the Viy and Vix since I know that the vertical and horizontal components must be equal (because it is 45 degress)?

Of course you can.
 

What is a "Difficult Projectile problem"?

A "Difficult Projectile problem" is a type of physics problem that involves calculating the trajectory, velocity, and other factors of a projectile in motion. These problems can be challenging because they often require the use of advanced mathematical equations and concepts.

What are some common examples of "Difficult Projectile problems"?

Some common examples of "Difficult Projectile problems" include calculating the trajectory of a bullet fired from a gun, determining the optimal angle for a soccer player to kick a ball into a goal, and finding the maximum height reached by a ball thrown into the air.

What are the key factors to consider when solving a "Difficult Projectile problem"?

The key factors to consider when solving a "Difficult Projectile problem" are the initial velocity of the projectile, the angle at which it is launched, the effects of gravity, and any other external forces acting on the projectile.

What are some tips for solving "Difficult Projectile problems"?

Some tips for solving "Difficult Projectile problems" include breaking the problem into smaller, more manageable parts, using equations that describe the motion of projectiles, and double-checking your calculations for accuracy.

Why are "Difficult Projectile problems" important in the field of science?

"Difficult Projectile problems" are important in the field of science because they help us understand and predict the motion of objects in the real world. They also allow us to design and optimize technologies such as rockets, missiles, and sports equipment.

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