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dxu

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Sorry if I ask a few too many questions, just a bit confused.

Man jumps over a bench at 7 m/s with angle of 50 respective to the horizontal. Find the total horiztonal distance.

First I broke it down to into Vx and Vy components for -1.8 and 6.7 respectively.

Then I solved for Vy = Vyi + gt which gave me 0.7
This was then multiplied by 2 since 0.7 is for only the first half of the arc. This gave me 1.4

I then solved for X using X = Xi + 1/2 (Vx + Vxi) t which gave me 1.26 m

Is this right?

Thanks
 

milski

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Your approach is correct but that's 50 degrees, not 50 radians. Having a negative component should have been a hint there. ;)
 

JRWPREMED

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Sorry if I ask a few too many questions, just a bit confused.

Man jumps over a bench at 7 m/s with angle of 50 respective to the horizontal. Find the total horiztonal distance.

First I broke it down to into Vx and Vy components for -1.8 and 6.7 respectively.

Then I solved for Vy = Vyi + gt which gave me 0.7
This was then multiplied by 2 since 0.7 is for only the first half of the arc. This gave me 1.4

I then solved for X using X = Xi + 1/2 (Vx + Vxi) t which gave me 1.26 m

Is this right?

Thanks


I don't think you are using the correct formula to find the component of Velocity vector.

You use cosd and sind to find the Vx and Vy components.

I'm getting Vx to be close to 4.5 m/s.

Rework the problem using 50 degree (Make sure your calc isn't in Radians), and you will get the correct answer. It honestly should have thrown you off when you got a negative Vy velocity. :p

Good Luck!
 

dxu

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Boy do I feel dumb. Thanks for the help guys, I got the answer. Amazing how something so innocuous can change the entire path. Note to self: pay the @#$% attention!!!
 

discowisco

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This might be a dumb question but isnt the horizontal velocity constant? I know in the case of projectile motion that the horizontal velocity is constant.

For this problem couldnt you just find the horizontal velocity using Vicos50 then do that * time (once you solve for it using Vy = Vyi + gt)
 
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milski

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This might be a dumb question but isnt the horizontal velocity constant? I know in the case of projectile motion that the horizontal velocity is constant.

For this problem couldnt you just find the horizontal velocity using Vicos50 then do that * time (once you solve for it using Vy = Vyi + gt)

Vx is constant and once you find the time, you have everything you need. He is doing it in a slightly more general way but that does not change the outcome. The real problem with the first solution is that he was calculating cos of 50 radians, not 50 degrees.
 

discowisco

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Thanks milski quick question though- how do you know for future problems when horizontal motion is constant? Just when there is nothing acting upon the horizontal direction right? I was second guessing my self whether the horizontal motion was constant in this case.

Also how do you know what sin 50 is ? I know the basics sin 0 30 60 90. I thought on the MCATs they include basically those angles??

Sorry for the basic questions
 

milski

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Like any other motion, the velocity will be constant if there are no forces acting. For a velocity component, like Vx, that means that there are no forces acting in that direction or in other words, all forces acting on the object are perpendicular to the direction of movement.

In the projectile case the only force acting is gravity and it is perpendicular to the Vx component. An example where this is not true is the period of time during which the projectile is being fired/accelerated - during that time there is an extra force acting on it which has component both in the x and y direction.

cos 50 is a bit less than cos 45 which is sort(2)/2 which I happen to know to be about .71. Nearest convenient number is 0.66 or 2/3 which should be a pretty good estimate for MCAT purposes.

One more point - sin 50 and cos 50 are both close to sin/cos 45 which are the same number. If you are getting significantly different Vx and Vy components, something is not right.
 
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