projectile..

[mak10]

Imagine a projectile being launched up a frictionless plane inclined at 60 degrees to the horizontal.... with a velocity of 60 m/s. How can you use this two data given to determine the distance travelled by the particle up the plane and the time taken for it do that??

 

I was quite confused by this since i get differing answers to distance using different methods. If I find t by t=u*sin 60/g... and plug the t value in s=u*sin 60*t - 1/2*9.8*t^2.... I get the distance as 137m. But my teacher uses the equation v^2 = u^2 - 2*9.8*sin 60*s taking v=0 and gets the distance as 212m. So in a nutshell, I am stuck... I got my exams tomorrow... so if any1 could solve this out for me... will be greatly appreciated. Am I right or is my teacher correct or none of us??

 

-mak10

[swansont]

Technically your teacher should be using cos30 for that equation, since it's a dot product, but you get the same answer.

 

Your coordinate system has to be consistent. Using the incline plane as the x-axis:

 

In your first equation, the acceleration is g*sin60, so the equation should be t = u/(g*sin60)

 

You shouldn't be using u*sin60 in the second equation, since the velocity in that direction is actually 60 m/s. The acceleration in that direction is -g*sin60, so that part is OK.

[cyeokpeng]

Hope I m still on time.

 

The fault is not the equation of motion's fault. It is correct whichever equation of motion you use, you will still get the right answer.

 

In the positive y-direction (vertical)

-g=(v-u)/t

v = u-gt

Since the final velocity in the y-direction is zero,

t = usin60/g

 

This is correct!

 

However, when you apply the displacement equation

s = u*sin60*t-0.5*g*t^2

You should remember that this is the vertical y displacement, NOT THE DISPLACEMENT UP THE RAMP!

since u*sin60 is your initial velocity in the vertical y-direction and your acceleration is your g.

 

To get the correct distance up the ramp, use trigonometry to find the ramp distance for this vertical distance 137.8m. I have a feeling that your teacher formula is wrong, unless you give me slightly wrong question.

COS

sin60 = 137.8/ramp distance

ramp distance = 137.8/sin60 = 159.1 m

 

Check your question again!

[mak10]

that makes sense. find the vertical distance first and then using trig. to find the distance up the plane... you do get 159.1m . But get this... if you find the horizontal distance as well by u*cos60*t and use pythagoroas theorem... you get a different answer... 210m. now, I am really confused!

[Severian]

Defining x as the distance up the ramp and calling the angle of the ramp [math]\theta[/math], you have:

 

The force in the x direction: [math]F=-mg \sin \theta=m a[/math]

Integrate this to get: [math]v=u-g \sin \theta t[/math]

Integrate again: [math]x=ut-\frac{g}{2} \sin \theta t^2[/math]

 

At the maximum, [math]v=0=u-g \sin \theta t[/math] so [math]t=\frac{u}{g \sin \theta}[/math] (this is what Swantson said)

 

Putting this into the equation for x: [math]x=\frac{u^2}{g \sin \theta} - \frac{u^2}{2g \sin \theta} = \frac{u^2}{2g \sin \theta} = \frac{60^2 m^2s^{-2}}{2 \times 9.81 ms^{-2} \sin 60^o} = 212m[/math]

[mak10]

oh well... guess my teacher was right. The problem was rather simple... the only difficulty I found was resolving g to find the acceleration up the ramp. But nonetheless... 210m too comes quite close to severian's value... so I think I can use either methods to get the answer. thanks for all your help guys!

 

-mak10

[cyeokpeng]

You should not apply it in the x-direction, since the acceleration is not g. Anyway, an easier method is to treat the ramp as the x-direction and use this coordinate system to solve the problem.