I was way over thinking this one.
It should just be released 10 ft before the porch because in 1 second after the paper is thrown it will travel 10 ft forward and 50 ft to the left (hitting the porch)
at least i think so haha..
A paperboy is riding at 10 ft/s on a bicycle and tosses a paper over his left shoulder at 50 ft/s. If the porch is 50 ft off the road, how far up the street should the paperboy release the paper to hit the porch?
So i drew a diagram that looks like a right triangle (attached)
I found the hypotenuse of the triangle to be
I figured if after he threw the paper at that speed then he would need to throw it before the porch.
It seems pretty far but i wasnt too sure about how else to go about this problem
Thank you for any help
Hi
I'm new member but i found the answer
on yahoo so here it is:
Call +y: direction of road, call +x: rightward direction
g: ground, b: bike, p: paper
Express relative motions as vectors. Neglect the vertical component of the paper's motion and assume it was the horizontal velocity that was given.
v_b:g = 10ft/s*<0,1>
v_p:b = 50 ft/s*<1,0>
Add them up to find velocity of paper:
v_p:g = v_b:g + v_p:b
v_p:g = <50, 10> ft/s
Express distance traveled of paper:
d = v_p:g * t
d = <50 * t, 10*t> ft/s
Set target:
d = <50 ft, x>
find x:
50ft = 50ft/s*t
t = 1sec
x = 10 ft
Conclusion: He should throw 10 ft uproad.