In Fig. 8-30, a 4.5 g ice flake is released from the edge of a hemispherical bowl whose radius r is 28 cm. The flake-bowl contact is frictionless. (a) What is the speed of the flake when it reaches the bottom of the bowl? (b) If we substituted a second flake with 4 times the mass, what would its speed be? (c) If, instead, we gave the flake an initial downward speed 1.4 m/s along the bowl, what would the answer be? I know for part a. mgh = 1/2mv^2 v = Sqaure root (2gh) v = 2.34 m/s part b i know is the same answer as a, because the mass does not effect the speed. but for part c, i am not getting the right answer and it seems like a relatively basic step. can someone help me out? Also, this problem is giving me headaches. Figure 8-34 shows a pendulum of length L = 1.4 m. Its bob (which effectively has all the mass) has speed v0 when the cord makes an angle 0 = 48° with the vertical. (a) What is the speed of the bob when it is in its lowest position if v0 = 9.9 m/s? What is the least value that v0 can have if the pendulum is to swing down and then up (b) to a horizontal position, and (c) to a vertical position with the cord remaining straight? http://edugen.wiley.com/edugen/courses/crs1062/art/qb/qu/c08/Fig08_34.gif i was trying to find the y-component. by doing cos48 = y/1.4 y = .94m then sin 48 = x/1.4 x = 1.04m 1.4 - .94 = 0.46m then i used v = Square root ((2)(9.8)(.46) v = 9.016m/s But then where do i use the initial velocity @? im just a bit confused. thanks for ur help!