Question

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?

Answer 1

So then he throws the ball straight up vertically?

Answer 2

i mean 40degrees woops

Answer 3

:-) Yeah I thought that was odd lol

Answer 4

aha yeah, couldn't read my notes for a second.

Answer 5

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.

Answer 6

yeah that's what makes it super hard. man i don't get how physics is so hard.

Answer 7

I think I'll just assume that initial y-position is y=0 and final y=position is also y=0.

Answer 8

*y-position

Answer 9

yeah @brinethery you are correct

Answer 10

is there any other information that i should know?

Answer 11

or assume*

Answer 12

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

Answer 13

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 :-(

Answer 14

hmm, well i'll check it over and compare it with my work, thanks though brinthery! :)

Answer 15

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.

Answer 16

@Algebraic! what would g stand for?

Answer 17

9.8, 9.81 or 10 m/s^2 depending on what your text prefers to use...

Answer 18

I left out a '1/2' on that last equation; should be:
\[5+\frac{ 1 }{ 2} a(\frac{16\sin40 }{g } )^2=8\cos40(\frac{ 16\sin40 }{g } )\]

Answer 19

questions about this @tetsirou ? or you good..?

Answer 20

@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.