How Do You Calculate Initial Velocity with Optimal Angle in Projectile Motion?

[fro]

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.

 

[neutrino]

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.

 

[radou]

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 $$x(t)=v_{0x}\cos(45)\cdot t$$. 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 $$y(t)=v_{0y}\sin(45)\cdot t-\frac{1}{2}\cdot g \cdot t^2$$. 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.)

 

[fro]

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.

 

[neutrino]

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. :)

 

[fro]

@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)?

 

[neutrino]

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.

 

[radou]

...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.