anonymous
  • anonymous
Angled Projectiles: A quarterback stands 5m behind a receiver. He throws the football at 8 m/s at an angle of 40° and wants the receiver to catch it. If the Receiver starts from rest and runs across level ground, what constant acceleration should he have to catch the ball? How far from the quarterback does the ball land?
Physics
schrodinger
  • schrodinger
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anonymous
  • anonymous
So then he throws the ball straight up vertically?
anonymous
  • anonymous
i mean 40degrees woops
anonymous
  • anonymous
:-) Yeah I thought that was odd lol

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anonymous
  • anonymous
aha yeah, couldn't read my notes for a second.
anonymous
  • anonymous
I hate the fact that we don't know the initial and final height of the ball because obviously one is holding it and one is catching it.
anonymous
  • anonymous
yeah that's what makes it super hard. man i don't get how physics is so hard.
anonymous
  • anonymous
I think I'll just assume that initial y-position is y=0 and final y=position is also y=0.
anonymous
  • anonymous
*y-position
anonymous
  • anonymous
yeah @brinethery you are correct
anonymous
  • anonymous
is there any other information that i should know?
anonymous
  • anonymous
or assume*
anonymous
  • anonymous
So the initial velocity has an x- and y-component because it's a vector. The x-component of velocity remains the same throughout the kick. If there was no ground for it to land on, it would continue on forever. The y-component of velocity obviously changes because we're dealing with acceleration due to gravity. So I'm going to look at the time for the ball to get to the top. At the top, final y-velocity is 0. So: Vi = 8sin(40), Vf = 0, a = -9.81, and we're looking for t V = Vi + at 0 = 8sin(40) -9.81(t) t = 0.52419 s This is the time it takes to get to the top. I could easily take a shortcut and double that time (because of symmetry), but I won't because you might have problems in the future that have different initial- and final y-positions. If you tried doubling it in that case, you'd end up with the wrong answers, so I'll take the long way for now. Okay so now we need to find the height at t=0.52419 s: y = y_o + v_o*t +.5at^2 y = 0 + (8sin(40))*(0.52419) + (0.5)(-9.81)(0.52419)^2 y = 1.35 m Now we can find the final y-velocity: v^2 = v_o^2 -2a(y-y_o) v^2 = 0 - 2(-9.81)(1.35-0) v = sqrt (26.487) = -5.147 m/s (minus because it's going down) Now we can find the time from y = 1.35 m to y=0 m (where it lands) y-y_o = ((v +v_o)/2)*t 1.35 - 0 = ((-5.147 +0)/2)*t t = 0.5251 s (I must have some rounding error here) So the total time it takes to get from the initial position to the final is 0.52419 s + 0.5251 s = 1.050s Now, we know the total time, we know that the x-component of the initial velocity remains constant. With this, we can calculate the total distance using the first equation: http://www.rdoman.com/phy/motf/motdrw01.gif (8cos(40))*(1.050s) = 6.435 m
anonymous
  • anonymous
Bump this question and see what someone else says in case I'm wrong. I get the concepts of projectile motion, but I tend to make mistakes :-(
anonymous
  • anonymous
hmm, well i'll check it over and compare it with my work, thanks though brinthery! :)
anonymous
  • anonymous
time in flight: \[0 = 8\sin40t - \frac{ 1 }{ 2}(g t^2)\] \[t= \frac{ 16\sin40 }{ g} \] receiver distance = distance ball travels: \[5 + \frac{ 1 }{ 2}(a t^2) = 8\cos40t\] plugging in time from the first part: \[5 + a( \frac{ 16\sin40 }{ g} )^2 = 8\cos40 (\frac{ 16\sin40 }{g })\] solve for a.
anonymous
  • anonymous
@Algebraic! what would g stand for?
anonymous
  • anonymous
9.8, 9.81 or 10 m/s^2 depending on what your text prefers to use...
anonymous
  • anonymous
I left out a '1/2' on that last equation; should be: \[5+\frac{ 1 }{ 2} a(\frac{16\sin40 }{g } )^2=8\cos40(\frac{ 16\sin40 }{g } )\]
anonymous
  • anonymous
questions about this @tetsirou ? or you good..?
anonymous
  • anonymous
@Algebraic! im trying to work on it, yet i'm still confused like what was your dx vx and all that stuff. the prequisites to solving the equation.

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