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anonymous
 3 years ago
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?
anonymous
 3 years ago
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?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So then he throws the ball straight up vertically?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i mean 40degrees woops

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0:) Yeah I thought that was odd lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0aha yeah, couldn't read my notes for a second.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0yeah that's what makes it super hard. man i don't get how physics is so hard.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think I'll just assume that initial yposition is y=0 and final y=position is also y=0.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah @brinethery you are correct

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is there any other information that i should know?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So the initial velocity has an x and ycomponent because it's a vector. The xcomponent of velocity remains the same throughout the kick. If there was no ground for it to land on, it would continue on forever. The ycomponent 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 yvelocity 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 ypositions. 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 yvelocity: v^2 = v_o^2 2a(yy_o) v^2 = 0  2(9.81)(1.350) 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) yy_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 xcomponent 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
 3 years ago
Best ResponseYou've already chosen the best response.0Bump 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
 3 years ago
Best ResponseYou've already chosen the best response.0hmm, well i'll check it over and compare it with my work, thanks though brinthery! :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0time 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
 3 years ago
Best ResponseYou've already chosen the best response.0@Algebraic! what would g stand for?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.09.8, 9.81 or 10 m/s^2 depending on what your text prefers to use...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0questions about this @tetsirou ? or you good..?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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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