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Projectile motion vector problem

  • #1

Homework Statement


A cannonball is fired from a point 35ft above ground level at an angle of 35 degrees. It hits the ground 100ft away. Calculate its travel time, initial speed, and maximum height above the ground.
initial height = h = 35ft
angle = x = 35 degrees
gravity = g = 32ft/s^2
travel time, initial speed, and max height are unknown.

Homework Equations


r(t)=(v*cosx)ti+[h+(v*sinx)t-(1/2)*g*t^2]j
initial velocity = sqrt[(Range*gravity)/sin(2x)]
Bolds represent vectors. r(t) = position function.

The Attempt at a Solution


I've found the derivative and attempted to find max height, but that didn't work without a velocity. I tried the initial velocity equation I posted but that only works if the initial height and final height are equal. I attempted to find the time and initial velocity by using r(t) = 100i + 0j making it so (v*cosx)ti = 100 and [h+(v*sinx)t-(1/2)gt^2]j = 0 But no matter what I did I would find myself with two unsolved variables.
 

Answers and Replies

  • #2
gabbagabbahey
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I I attempted to find the time and initial velocity by using r(t) = 100i + 0j making it so (v*cosx)ti = 100 and [h+(v*sinx)t-(1/2)gt^2]j = 0 But no matter what I did I would find myself with two unsolved variables.

You have two equations; [itex]v\cos(x)t=100[/itex] and [itex]h+v\sin(x)t-\frac{1}{2}gt^2=0[/itex], and two unknowns; [itex]v[/itex] and [itex]t[/itex]. You should be able to solve for both [itex]v[/itex] and [itex]t[/itex] easily. So, if you are having difficulty, post your calculations.
 
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  • #3
cos(35) = 0.82, sin(35) = 0.57
x = 0.82(v)(t)
y = 35 + 0.57(v)(t)-16t^2
0 = 0.82(v)(t)
0 = 0.57(v)(t) - 16t^2
0.82(v)(t) = 0.57(v)(t) - 16t^2
16t^2 - 0.25(v)(t) = 0
At which point I'm stuck, because I would complete the square, but the v gets in the way of that.

or

v(t) = 0.82vi + (0.57v - 32t)j
0.57v - 32t = 0
t = 0.018v
Would give me the max height, at which point I could simply double t to find the Range to use the formula I posted earlier, but once again, the v gets in the way. I would end up with
x = .03v^2
y = .02v^2 - .02v^2 = 0

And I can't find any way to put that to use.
 
  • #4
gabbagabbahey
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Both equations must be true simultaneously. Start by isolating [itex]v[/itex] in the first equation, then substitute that into the second equation and solve it.

This process of solving two simultaneous equations is something you should have been taught in high school algebra.
 
  • #5
I do know that method, and maybe I'm just being really dense for some reason, but I'm just not seeing how I can properly apply it here. I've been looking over this problem for the last few hours, and I'm just not seeing how I can connect these equations and make this work.

Part of the problem I think is that I only have x and y values for t = 0 at which point velocity doesn't come into play. I think I could manage if I had x and y values for a different value of t, but as you can see in the problem write up I don't.
 
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  • #6
gabbagabbahey
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I do know that method, and maybe I'm just being really dense for some reason, but I'm just not seeing how I can properly apply it here. I've been looking over this problem for the last few hours, and I'm just not seeing how I can connect these equations and make this work.

Part of the problem I think is that I only have x and y values for t = 0 at which point velocity doesn't come into play. I think I could manage if I had x and y values for a different value of t, but as you can see in the problem write up I don't.
I'm not sure how I can make this any clearer without just solving it for you.

You have two equations; [itex]v\cos(x)t=100[/itex] and [itex]h+v\sin(x)t-\frac{1}{2}gt^2=0[/itex], and two unknowns; [itex]v[/itex] and [itex]t[/itex].

Isolate [itex]v[/itex] in the first equation...what do you get?...Plug that into the second equation...what do you get?
 
  • #7
Thanks a lot for the help. I finally saw what you meant, ironically while going through it again on my own. So it turns out I was just being extremely dense, but thank you for pointing out the obvious for me. It seems that was just what I needed.
 

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