At a county fair, a boy and two friends bring their teddy bears on the giant Ferris wheel. The wheel has a diameter of 10.5 m, the bottom of the wheel is 1.7 m above the ground and its rim is moving at a speed of 1.0 m/s. The boys are seated in positions 45° from each other. When the wheel brings the second boy to the maximum height, they all drop their stuffed animals. How far apart will the three teddy bears land? (Assume that the boy on his way down drops bear 1, and the boy on his way up drops bear 3.)
A. Find distance between bears 1 and 2
B. Find distance between bears 2 and 3
D = v_0*t +(1/2)at^2
v_f^2 = v_i^2 + 2ad
The Attempt at a Solution
I was able to figure out part a, with relatively no problems. The answer is 3.163 m.
Part B seems to not be working for me.
I know that the coordinates for Bear 2 are (0, 12.2) with V_x_i = 1.0 m/s. It has no initial y velocity.
For bear 2:
-12.2 m = (1/2)(-9.81)t^2
t = 1.5771s
d = (1.5771 s)(1 m/s)
d = 1.5771 m from it's initial distance of 0
For Bear 2, because it's 45 degrees away, it's coordinates are (+2.625 m, 9.575 m). It's initial velocities are v_x = 1cos(45) = 0.707 m/s. v_y = -sin(45) = -.707 m/s
d_y = v_0_y*t + (1/2)(g)(t^2)
0 = -4.905*t^2 + -0.707*t + 9.575
t = 1.3269 s
d_x = (0.707 m/s)(1.32696 s)
d_x = 0.93816
Total d_x of bear 1 is 2.625 + 0.93816 = 3.56316 m
distance between bear 2 and 1 is 3.56316 - 1.5771 m = 1.9861 m, but that's wrong. Can anyone tell me what I'm doing wrong?
Thanks in advance!