1. A.P. Physics B
6.5 #’s 2-4
By: Anna Findley

2. 6.5 # 2
2. [CJ5 6.P.036.] A 47.0 g golf ball is driven from the tee with an initial speed of 52.0 m/s and rises to a height of 24.6 m.
(a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.
(b) What is its speed when it is 8.0 m below its highest point?

3. Solution Part A
(a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.
KE + mgh=½ mv2
KE+ (.047kg) (9.8) (24.6m) = ½ (.047kg) (52.0m/s) 2
KE= 52.2 J

4. Solution Part B
(b) What is its speed when it is 8.0 m below its highest point?
KE + mgh = ½mv2 + mg (h-8.0)
(52.2J) + (.047kg) (9.8) (24.6m) = ½(.047kg) v 2 + (.047kg) (9.8) (16.6m)
V= 48.77m/s

5. 6.5 # 3
3. [CJ5 6.P.037.] A cyclist approaches the bottom of a gradual hill at a speed of 11 m/s. The hill is 5.0 m high, and the cyclist estimates that she is going fast enough to coast up and over it without peddling. Ignoring air resistance and friction, find the speed at which the cyclist crests the hill.

6. Solution #3 KE + PE = KE + PE ½ mv2+ mgh = ½ mv2+ mg
V = 4.8 m/s
Remember: mass cancels out...

7. 6.5 # 4
*4. [CJ5 6.P.041.] A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. As the drawing shows, one person hits the water 5.00 m from the end of the slide in a time of s after leaving the slide. Ignoring friction and air resistance, find the height H in the drawing.

8. Solution # 4 Vx= d/t Vx= (5.00m) / (0.500s) Vx= 10.0 m/s
Vfy = Viy + at
Vfy = 0 + (9.8) (0.500s)
Vfy = (4.9m/s)

9. Solution # 4 Continued Use the Pythagorean Theorem to find Vf
Vx2 + Vfy2 = vf2
Vf = 11.14
Vf2= 2ad
= 2 (9.8) d
D = 6.33