Projectile Motion, missing variables.

In summary, Mike is having trouble with projectile motion. He does fine when he has all the variables, but when problems arise such as taking velocity or angle away, he gets stuck. He asks for help from someone who can help him out with the problem.
  • #1
Visual1Up
12
0
I am having some trouble with projectile motion, I do fine when I have all of the variables such as initial velocity, and angle, and say I want to find the range. That's great, I use y = voyt - 1/2 gt^2 where y = 0 to find the time, since voy = vo sin theta, and r = voxt where vox = vo cos theta. But I am having trouble when problems are taking the velocity or angle away.

Such as...theta is 35, range is 4m, find the initial velocity... something along those lines. I get horribly stuck because now I "can't" find voy, or vox, etc. Would anyone mind helping? Thanks!,

-Mike :bugeye:
 
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  • #2
You'd tackle this one by first considering horizontal motion. You don't know Vo but you kwo the relationship between Vo, theta and t because you have the distance.

So you have an expression for t in terms of v (t = 4/vcos35)

Then you look at the vertical remembering that when the stone has landed, the vertical displacemetn is zero

y = voyt - 1/2 gt^2 using the expression you have for t.

You end up with v as the only variable.

Does that help?
 
  • #3
Visual1Up said:
Such as...theta is 35, range is 4m, find the initial velocity... something along those lines. I get horribly stuck because now I "can't" find voy, or vox, etc. Would anyone mind helping? Thanks!,

-Mike :bugeye:

It's all about doing a little backwards thinking. :smile:
 
  • #4
rsk said:
You'd tackle this one by first considering horizontal motion. You don't know Vo but you kwo the relationship between Vo, theta and t because you have the distance.

So you have an expression for t in terms of v (t = 4/vcos35)

Then you look at the vertical remembering that when the stone has landed, the vertical displacemetn is zero

y = voyt - 1/2 gt^2 using the expression you have for t.

You end up with v as the only variable.

Does that help?

I think I understand, your saying the equation turns into y = 1/2gt^2? since voy = 0. but when working that I get 42 m/s and t = .1 . That can't be right. :rolleyes: . Excuse me if I am missing something, I am having a hard time with physics.
 
  • #5
Visual1Up said:
I think I understand, your saying the equation turns into y = 1/2gt^2? since voy = 0.

No - it has vertical velocity. But when it lands it has no vertical displacement, so y = 0, not Voy

Voy = Vo sin theta
 
  • #6
oh, so are you saying... 0 = v sin 35 - (1/2)g * (4/(v sin 35))^2 ?

EDIT: v cos 35 on the right
 
  • #7
0 = Vosin35 t - 1/2 g t^2

0 = Vosin35 4/cos35 - 1/2 g (4/Vocos35)^2

I think you missed the t from the first part.
 
  • #8
Visual1Up, I think you know how to derive the formula that gives range in function of [itex]\theta , v_o[/itex], which is:

[itex]R = (v_0^2 \sin 2\theta)/g[/itex]

Then, it's easy. :smile:
 
  • #9
Well with rsk's problem, I can't ever arrive at the right answer because the math is killing me. I have never seen zeno's problem before but when working I get vo = .649 and time = 7.5 sec which can't be right. I am horrible at this :(
 
  • #10
Ok, go from here

0 = Vosin35 t - 1/2 g t^2

and let's cancel a t from each part before we start...

0 = Vosin35 - 1/2 g t

Subs in for t, from before

0 = Vosin35 - 1/2 g (4/Vocos35)

Or in other words Vosin35 = 1/2 g (4/Vocos35)

Rearranging gives

Vo^2 = 1/2 g 4 /(sin35 cos35)
Work out your sines and cosines and you should get a value of 6.46 for Vo

I've worked it backwards to find t, and to check the x and y displacements, and it works.
 
  • #11
rsk said:
Ok, go from here

0 = Vosin35 t - 1/2 g t^2

and let's cancel a t from each part before we start...

0 = Vosin35 - 1/2 g t

Subs in for t, from before

0 = Vosin35 - 1/2 g (4/Vocos35)

Or in other words


Vosin35 = 1/2 g (4/Vocos35)

Rearranging gives

Vo^2 = 1/2 g 4 /(sin35 cos35)



Work out your sines and cosines and you should get a value of 6.46 for Vo

I've worked it backwards to find t, and to check the x and y displacements, and it works.

ahhhh I forgot to factor out a t :frown: ok I got 6.46. It seems like the math does me in more than the formula's. But thank you very much for the help.
 

Related to Projectile Motion, missing variables.

1. What is projectile motion?

Projectile motion is the motion of an object through the air that is affected only by the force of gravity. This type of motion is commonly seen in objects that are thrown, launched, or dropped.

2. What are the variables involved in projectile motion?

The variables involved in projectile motion are initial velocity, angle of launch, time, displacement, and acceleration due to gravity. These variables are used to calculate the trajectory and final position of the object.

3. How do I find missing variables in projectile motion problems?

To find missing variables in projectile motion problems, you can use the equations of motion, such as the kinematic equations, to solve for the unknown variable. It is important to set up the problem carefully and use the appropriate equation for the given scenario.

4. What are some real-life examples of projectile motion?

Some real-life examples of projectile motion include throwing a ball, jumping off a diving board, shooting a basketball, and launching a rocket into space. These are all situations where an object is in motion due to the force of gravity.

5. How does air resistance affect projectile motion?

Air resistance can affect projectile motion by slowing down the object and changing its trajectory. This is especially noticeable in situations where the object has a large surface area, such as a parachute or a feather, and experiences significant air resistance. In these cases, the object may not follow a perfect parabolic path and will have a shorter horizontal range.

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