
#1
Sep2305, 02:06 PM

P: 1

I have several vector problems that I really need help with and I have a test on Monday. I am really screw.
1) By throwing a ball at an angle of 45 degree, a girl can throw it a maximum horizontal distance of R= 33 m on a level field. How high can she throw the same ball vertically upward? Assume that her muscles give the ball the same speed in each case. This is what I have done so far, and I do not wheter I did my work right or not. displacement x= 33 m, t=?, Vi= ? 33/(Vi*(cos(45)))=t Then I substitute t in y=Vi(33/(Vi*(cos(45)))+(4.9)(33/(Vi*(cos(45)))^2, and I got t=8.87 sec. And I am stuck. Thanks for your help 



#2
Sep2305, 02:19 PM

P: 390

Well, you found the total time, which means at that point in time y = 0, so you set that equation = 0
Also, I think you mean Visin45 * 33/.... since that's the vertical velocity. In the end you'll end up with a dirty quadratic equation in Vi. :) 



#3
Sep2305, 04:41 PM

P: 64

Im gonna have to disagree with the both of you. though solving the equation using a time parameter will eventually give you the answer, the math is messy and overly complicated. for this problem, you weren't given time and weren't asked to solve for the time. It seems that the time variable is not only not neccesary, but not the "optimal choice"
The two equations you wanna work with here is the horizontal range equation and the alternate kinematic equation of velocity. the range equation is: R=v^2sin2(theita)/g (V= initial velocity,not horizontal) Vy^2=Vy0^22g*delta y Since you were given the range and the launch angle, you can use the first equation to solve for the initial velocity (g=9.8 in case anyone forgets). From that, you can figure out the delta y since you know the Vy0 and since Vy is zero at the top of its flight. Sorry for the information and verbal overload, I just though this setup would be slightly more elegant and simple to solve. 



#4
Sep2305, 04:58 PM

P: 390

I am stuck with vector problems
The R = v^2sin blahblah formula you're giving us is a derived formula. :P
If I was solving for the range using the equations we used, I would end up with the formula you have there. :P 



#5
Sep2305, 05:39 PM

P: 789

From there you can solve for [tex]\vec{V_{0}}[/tex] using the kinematic equation you showed. 



#6
Sep2305, 06:30 PM

Sci Advisor
HW Helper
P: 2,275

Actually, Katrina is about halfway there. Now just divide the horizontal distance by the time to find the initial velocity (33/8.87) At that point, your idea of using half the time would be a good idea. You know the object will slow down on the way up at the same rate it will speed up on the way down. It will reach its peak at the halfway point. You redo the [tex]s=v_i t + \frac{1}{2} a t^2[/tex] using half the time instead of the time to reach the ground again. Edit: Oops, my bad. I just assumed that since you set up the equation, you'd wind up with the right answer for t. You made a mistake somewhere along the line because t is too big. In fact, with your equation set up the way it is, you'll find initial velocity before you find time. By the way, you don't have to solve a quadratic equation with your equation set up the way it is. 



#7
Sep2305, 06:46 PM

P: 789

I wasn't saying that the two speeds are equal, but that you can solve for the height by using trigonometry. To solve for [tex]\vec{v_0}[/tex] I was saying to use the equation that DaMastaofFisix gave: [tex]v_{yf}^2=v_{y0}^2+2gy[/tex]. And you can use that to find the initial velocity.
I think that's correct. It's been a long week. 



#8
Sep2305, 11:13 PM

HW Helper
PF Gold
P: 1,123

The appropriate way to do this is to solve for t to hit the ground
(vi sin45) = gt/2 , so the vi^2 don't cancel. set the distance scale with the horizontal motion. Once you get the correct vi^2/g , most folks would use (v_f)^2 = (v_i)^2 + 2 a.x (which is generally valid, even if it looks like a "math trick"). Finally, explain why there are two sqrt(2) factors. 


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