Grade 11 Intro to Physics question - distance a football travels

[FirstPhysics]

Homework Statement

How far will a football travel when it is thrown at 62 km/h at an angle of 35° to the ground?

Homework Equations

V1v = v1 sinθ; Δd= V1v * Δt + 1/2 * (-9.8 m/s^2) * Δt^2;

The Attempt at a Solution

V1v = v1 sinθ
V1v = 17.2 m/s sin35° [up]
=9.9 m/s [up]

Δd= = V1v * Δt + 1/2 * (-9.8 m/s^2) * Δt^2 (Factor out Δt)
= Δt (9.9 m/s - 4.9 m/s^2 *Δt)
Δt = 0 or 2.0s

I do correspondence schoolwork and this question is just a support question and I won't be marked on it. The thing is I still need to know how to do the question (obviously). Here is where I run into my problem: I know how to figure everything out, EXCEPT for how they managed to get Δt = 0 or 2.0s. I don't know how to figure out the time.
Any help would be appreciated. And please dumb it down as much as possible - physics is not my strongest subject.

 

[LawrenceC]

You have the equation:

0 = Δt (9.9 m/s - 4.9 m/s^2 *Δt)
The 0 and 2 are the two roots to the quadradic equation. Which ever one you plug in, the result is zero. It is telling you that at time 0, the ball is on the ground - has not been tossed. The 2 seconds is the time when it returns to the ground. You can also solve the equation by using the formula for quadradics.

If you don't round off until the end of the computation, you'll get something very close to 2 seconds.

 

[BTe]

Split the question into 2 right angle triangles with the center being the highest point /|\

Figure out x and y components of the velocity, and with that use related formulas to figure out d, t, etc.

 

[yesgirl10]

You have the equation:

0 = Δt (9.9 m/s - 4.9 m/s^2 *Δt)
The 0 and 2 are the two roots to the quadradic equation. Which ever one you plug in, the result is zero. It is telling you that at time 0, the ball is on the ground - has not been tossed. The 2 seconds is the time when it returns to the ground. You can also solve the equation by using the formula for quadradics.

If you don't round off until the end of the computation, you'll get something very close to 2 seconds.

Hi! I'm working on this problem at the moment and don't exactly understand the time part either. Am I just supposed to plug numbers into the equation for Δt until it is equal to zero? Because when I tried putting in the 2.0s I just get 0.2 as an answer. Please help, I'm incredibly confused!

 

[Rellek]

Hey there!

This question has an actual equation behind it for the range of a trajectory, but I think they want you to derive it with this question.

So, I think you should figure out how long it takes for the football to hit the maximum height of it's trajectory. This equation should help out:

Vf = V0 - gt

(Remember to use the y component of this velocity.)

So, once you have this time, how long will it take to go from the maximum height back down to the ground? Well, since the arc of it's trajectory is symmetrical, the time to get to the peak height will also be equal to the time from the peak height back down to the ground.

So the total time is 2 times the time you calculated from the kinematic equation.

Now, for the horizontal component, notice that there is NO acceleration at all. So, the total horizontal distance of the trajectory will be D = V0*t (remember V = d/t)

So, what would you plug into this equation for the time component? The total time it is in the air. perhaps?

After that, it is just a matter of recognizing the double angle identity and you'll have an equation for your trajectory!

Hope this helps! Ask if you need clarification!

 

[yesgirl10]

Hey there!

This question has an actual equation behind it for the range of a trajectory, but I think they want you to derive it with this question.

So, I think you should figure out how long it takes for the football to hit the maximum height of it's trajectory. This equation should help out:

Vf = V0 - gt

(Remember to use the y component of this velocity.)

So, once you have this time, how long will it take to go from the maximum height back down to the ground? Well, since the arc of it's trajectory is symmetrical, the time to get to the peak height will also be equal to the time from the peak height back down to the ground.

So the total time is 2 times the time you calculated from the kinematic equation.

Now, for the horizontal component, notice that there is NO acceleration at all. So, the total horizontal distance of the trajectory will be D = V0*t (remember V = d/t)

So, what would you plug into this equation for the time component? The total time it is in the air. perhaps?

After that, it is just a matter of recognizing the double angle identity and you'll have an equation for your trajectory!

Hope this helps! Ask if you need clarification!

Wow, thank you so much for such a quick and thorough reply!

Soo, what I did was rearranged the equation to t = Vf-V0/g. Although I still can't really wrap my head around how this all works, it just does and is super helpful. So I did 9.9m/s [up] - 0m/s / 9.8m/s^2 [up] and then got approximately 1s. I proceeded to double it and got the 2s! Yay! haha.
And then to get the distance it traveled I multiplied the horizontal part, 14.1m/s [forward] by the 2s. And got the correct answer. :)

Thanks again, this helped a lot!

 

[yesgirl10]

Hey there!

This question has an actual equation behind it for the range of a trajectory, but I think they want you to derive it with this question.

So, I think you should figure out how long it takes for the football to hit the maximum height of it's trajectory. This equation should help out:

Vf = V0 - gt

(Remember to use the y component of this velocity.)

So, once you have this time, how long will it take to go from the maximum height back down to the ground? Well, since the arc of it's trajectory is symmetrical, the time to get to the peak height will also be equal to the time from the peak height back down to the ground.

So the total time is 2 times the time you calculated from the kinematic equation.

Now, for the horizontal component, notice that there is NO acceleration at all. So, the total horizontal distance of the trajectory will be D = V0*t (remember V = d/t)

So, what would you plug into this equation for the time component? The total time it is in the air. perhaps?

After that, it is just a matter of recognizing the double angle identity and you'll have an equation for your trajectory!

Hope this helps! Ask if you need clarification!

Wow, thank you so much for such a quick and thorough reply!

Soo, what I did was rearranged the equation to t = Vf-V0/g. Although I still can't really wrap my head around how this all works, it just does and is super helpful. So I did 9.9m/s [up] - 0m/s / 9.8m/s^2 [up] and then got approximately 1s. I proceeded to double it and got the 2s! Yay! haha.
And then to get the distance it traveled I multiplied the horizontal part, 14.1m/s [forward] by the 2s. And got the correct answer. :)

Thanks again, this helped a lot!

 

[yesgirl10]

Hey there!

I also have another question that I got confused about, so if you could help me out again please it would be greatly appreciated!

The question is:

A water balloon is fired at 34.0 m/s from a water cannon, which is aimed at an angle of 18 degrees above the ground. The center of the cannon's target (which has a radius of 1.0 m) is painted on the asphalt 42.0 m away from the water cannon.

a) Will the balloon hit the target? Justify your response with calculations that indicate where the water balloon will land.
b) Make one suggestion about how to adjust the water cannon so that the water balloon will hit the target. Justify your choice.

So for a) to get time I did Δt=10.5/9.8 and then multiplied that by 2 giving me 2.1s. To get the distance the balloon traveled I multiplied the 2.1s by the horizontal thing (velocity?) 32.3m/s [forward]. Therefore I got a total distance of about 67.8m. So obviously it does not hit the target and goes like 25.8m past it. So I think I have this part all down pat but please correct me if I've made any mistakes.

Part b) is where I got confused. I want to find what the launch angle has to be for the water balloon to hit the target at 42m. So, I tried using the equation:

Dh=V1^2*sin2θ/a --> rearranged: 2θ=arcsin(Dh*a/V1^2)

When I plugged in the values, 2θ=arcsin(42m*9.8m/s^2/34^2) I got that θ=10.43°.

I think this is wrong because I thought the angle had to get bigger in order for the horizontal distance to decrease? I don't know.
Could you please break this step down for me?

 

[Rellek]

I also have another question that I got confused about, so if you could help me out again please it would be greatly appreciated!

The question is:

A water balloon is fired at 34.0 m/s from a water cannon, which is aimed at an angle of 18 degrees above the ground. The center of the cannon's target (which has a radius of 1.0 m) is painted on the asphalt 42.0 m away from the water cannon.

a) Will the balloon hit the target? Justify your response with calculations that indicate where the water balloon will land.
b) Make one suggestion about how to adjust the water cannon so that the water balloon will hit the target. Justify your choice.

So for a) to get time I did Δt=10.5/9.8 and then multiplied that by 2 giving me 2.1s. To get the distance the balloon traveled I multiplied the 2.1s by the horizontal thing (velocity?) 32.3m/s [forward]. Therefore I got a total distance of about 67.8m. So obviously it does not hit the target and goes like 25.8m past it. So I think I have this part all down pat but please correct me if I've made any mistakes.

Part b) is where I got confused. I want to find what the launch angle has to be for the water balloon to hit the target at 42m. So, I tried using the equation:

Dh=V1^2*sin2θ/a --> rearranged: 2θ=arcsin(Dh*a/V1^2)

When I plugged in the values, 2θ=arcsin(42m*9.8m/s^2/34^2) I got that θ=10.43°.

I think this is wrong because I thought the angle had to get bigger in order for the horizontal distance to decrease? I don't know.
Could you please break this step down for me?

You are correct on all counts. Think about it ... if your object is overshooting the gap, then lowering the angle WOULD add a greater horizontal component to the velocity, but it would reduce the time in the air by a larger amount as well. Also note that 90° - θ (the 10° you found) would actually yield the same trajectory.

 

[Rellek]

Dh=V1^2*sin2θ/a --> rearranged: 2θ=arcsin(Dh*a/V1^2)

When I plugged in the values, 2θ=arcsin(42m*9.8m/s^2/34^2) I got that θ=10.43°.

I think this is wrong because I thought the angle had to get bigger in order for the horizontal distance to decrease? I don't know.
Could you please break this step down for me?

Also, a rather useful (and probably intuitive) feature that arises from this equation is the angle for maximal trajectory.

You actually wouldn't need to use calculus to prove this, you could just recognize that the maximum value of your sine function would be 1, and find which angle would cause that, but I think this way is more fun :)

So, take dr/dθ = 0 (the derivative of this function with respect to θ), and this means this would be the maximal possible trajectory.

dr/dθ = V0²*2cos2θ/g = 0, notice g, 2, and V0 will cancel out.

Your maximal θ will then occur AT cos2θ = 0, or 2θ = cos^-10,

Or, 2θ = 90, which shows that the angle of maximum possible trajectory will be 45°, in every case.

 

[yesgirl10]

You are correct on all counts. Think about it ... if your object is overshooting the gap, then lowering the angle WOULD add a greater horizontal component to the velocity, but it would reduce the time in the air by a larger amount as well.

Of course! This makes so much sense now, I had completely forgotten to take the actual time into account... Silly me. :tongue:

Also note that 90° - θ (the 10° you found) would actually yield the same trajectory.

Oh I didn't know that, so 79.57° works as well. But then since the angle is bigger it would take a shorter horizontal velocity and a much greater amount of time to hit the target right? So will there always be two answers/angles that work then? And to get the second one I just need to subtract my original one (in this case 10.43°) from 90°?

Or, 2θ = 90, which shows that the angle of maximum possible trajectory will be 45°, in every case

This part confused me a bit, so does this mean that 79.57° is not okay?

 

[Rellek]

This part confused me a bit, so does this mean that 79.57° is not okay?

No, both angles are the same, and you can use both angles. 45° will always yield the longest possible trajectory. Any angle under that will be less, and any angle above that will be less. Because of the symmetry of the trigonometric functions though, it just so happens that the same trajectory can occur with two different angles, and that relationship will be 90°-θ, once you've found one of the angles.

 

[yesgirl10]

No, both angles are the same, and you can use both angles. 45° will always yield the longest possible trajectory. Any angle under that will be less, and any angle above that will be less. Because of the symmetry of the trigonometric functions though, it just so happens that the same trajectory can occur with two different angles, and that relationship will be 90°-θ, once you've found one of the angles.

Gotcha, thanks a bunch! :)
Is it ok for future questions if I just send you a message?