calculating the verticle vector of this

[carldd]

calculating the verticle vector (speed) of this

Hi!

I have this problem.

A man has to kick a ball that lies on the ground. He has to get it to land exactly five metres away. The horizontal vector of the kick is known, but not the vertical vector. Lets say the horizontal speed is 7m/s (or whatever you like). How do I calculate the vertical speed supposing that theres no air resistance?:?:

I havent been able to find a solution on the Internet. Only if the combination of horizontal and vertical component is known, but thats different... Also this is like doing it all backwards, because in most examples the length of the throw is yet to be solved. In this example it's already given.

 

[wjm11]

Hi!

I have this problem.

A man has to kick a ball that lies on the ground. He has to get it to land exactly five metres away. The horizontal vector of the kick is known, but not the vertical vector. Lets say the horizontal speed is 7m/s (or whatever you like). How do I calculate the vertical speed supposing that theres no air resistance?:?:

I havent been able to find a solution on the Internet. Only if the combination of horizontal and vertical component is known, but thats different... Also this is like doing it all backwards, because in most examples the length of the throw is yet to be solved. In this example it's already given.

Try to figure out what you do know (horizontal velocity in this case) and what other things you can determine from that. In problems like this, horizontal velocity will remain constant. That is important to remember. That means we can use a simple d = vt formula for the horizontal displacement. So, in examining the horizontal portion of the problem, we already know both v (7 m/s) and d (5 m). That means we can easily calculate time, t = d/v = (5 m)/(7 m/s) = (5/7) sec.

Now use that info in the vertical part of the problem.

Your turn.

 

[carldd]

Try to figure out what you do know (horizontal velocity in this case) and what other things you can determine from that. In problems like this, horizontal velocity will remain constant. That is important to remember. That means we can use a simple d = vt formula for the horizontal displacement. So, in examining the horizontal portion of the problem, we already know both v (7 m/s) and d (5 m). That means we can easily calculate time, t = d/v = (5 m)/(7 m/s) = (5/7) sec.

Now use that info in the vertical part of the problem.

Your turn.

Ok. I dont know if this is correct but I give it a try. It was a long time a go I did this kind of math but...

From what I remember the acceleration of a free fall is linear. This is also true for a parabola caused by a throw a kick as in this case. No matter what angle I throw at, the falling speed is going to be constant. Because this is for a computer game , the gravity is not 9.8 but 6.82, so I need to calculate how far I can add 6.82^2 to the time of 5/7 seconds. This give me the equation
6.82y²=5/7 and solving for y: 6.82y=√

URL: http://able2know.org/topic/20190-1



URL: http://able2know.org/topic/20190-1
​

(5/7) 6.82y=0.845. y=6.82/0.845=8.07

Is that correct?

 

[carldd]

Try to figure out what you do know (horizontal velocity in this case) and what other things you can determine from that. In problems like this, horizontal velocity will remain constant. That is important to remember. That means we can use a simple d = vt formula for the horizontal displacement. So, in examining the horizontal portion of the problem, we already know both v (7 m/s) and d (5 m). That means we can easily calculate time, t = d/v = (5 m)/(7 m/s) = (5/7) sec.

Now use that info in the vertical part of the problem.

Your turn.

Thanks.

Yeah because the acceleration is linear no matter what angle I shoot the ball I think this equation will do:

9.81² * x = 5/7 Because this is for a computer game, the computer has to solve for x. Im not very good at algebra so how can I define x from that equation without solving for x? (I know hot to solve for x, but I need to get x on the left side alone).

EDIT: Never mind I found out

 

[carldd]

Oh and im not very good with algebra but I think this is right? I guess that should be x=(5/7) / 9.81².

EDIT: Im not sure the equation is correct anymore because I cant get it to work. Ive been working by this two weekends now and Im stuck... Can you please tell me if its incorrect? What I ask for is a formula that calulates the vertical force, only by knowing the horizontal force. So I can put it inside a script for my game.

YEAH!!! I Solved it at least.

I worked out a formula its the air time squared * half of the gravity (9.81*0.5). At least I think so

 

[DrPhil]

A man has to kick a ball that lies on the ground. He has to get it to land exactly five metres away. The horizontal vector of the kick is known, but not the vertical vector. Lets say the horizontal speed is 7m/s (or whatever you like). How do I calculate the vertical speed supposing that theres no air resistance?

Oh and im not very good with algebra but I think this is right? I guess that should be x=(5/7) / 9.81².

EDIT: Im not sure the equation is correct anymore because I cant get it to work. Ive been working by this two weekends now and Im stuck... Can you please tell me if its incorrect? What I ask for is a formula that calulates the vertical force, only by knowing the horizontal force. So I can put it inside a script for my game.

There are two parameters, angle and velocity. (\(\displaystyle \theta,\ \ v\)).

The horizontal component is \(\displaystyle v_H = v\ \cos\theta\), which is constant. Thus the time it takes to go 5m in the horizontal is

\(\displaystyle T = d/v_H = \dfrac{(5.0\ \mathrm m)}{v\ \cos\theta}\)

In that same time interval, the ball will rise to some height and return to the ground. It has an initial velocity \(\displaystyle v_0 = v\ \sin\theta\), and a downward acceleration of \(\displaystyle g = 9.81\ \text{m/s}^2\). The vertical equation of motion is

\(\displaystyle y(t) = v\ \sin\theta\ t - \frac 12 (9.81\ \text{m/s}^2)\ t^2\)

Specifically, to return to the ground at the same instant it has gone 5.0 m horizontally, we must have

\(\displaystyle y(T) = 0 = v\ \sin\theta\ T - \frac 12 (9.81\ \text{m/s}^2)\ T^2\)
.............\(\displaystyle = v\ \sin\theta\ \dfrac{(5.0\ \mathrm m)}{v\ \cos\theta} - (4.905\ \text{m/s}^2)\ \left(\dfrac{(5.0\ \mathrm m)}{v\ \cos\theta}\right)^2\)

\(\displaystyle v\ \sin\theta = (4.905\ \text{m/s}^2)\ \dfrac{(5.0\ \mathrm m)}{v\ \cos\theta}\)

After applying the double-angle formula \(\displaystyle \sin(2\theta) = 2\ \sin\theta\ \cos\theta\), the final result is

\(\displaystyle v^2\ \sin(2\theta) = 49.05\ (\text{m/s})^2\)......or......\(\displaystyle v\ \sqrt{\sin(2\theta)} = 7.0 \text{ m/s}\)