{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# S07s(3) - Physics 204A FINAL EXAM Chapters 1 14 Spring 2007...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Physics 204A FINAL EXAM Chapters 1 - 14 Spring 2007 1 Name:__________________________________ Check here to have your grade posted on the class web site. Solve the following problems in the space provided. Use the back of the page if needed. Each problem is worth 10 points. You must show your work in a logical fashion starting with the correctly applied physical principles. The equations you need are on the equation sheet. Your score will be maximized if your work is easy to follow because partial credit will be awarded. 1. The pitcher throws a pitch from the mound toward home plate 18.4m away. The ball is released horizontally from a height of 2.00m with a velocity of 42.0m/s. Find (a)the time it takes for the ball to reach home plate and (b)the height above the ground when it gets there. (a)Use the kinematic equation along the x-direction to get the time, x = x o + v ox t + 1 2 a x t 2 . Plugging in the things that are zero and solving for t, x = v ox t " t = x v ox = 18.4 42.0 " t = 0.438 s . (a)Use the kinematic equation without v y to get the final height, y = y o + v oy t + 1 2 a y t 2 . The initial velocity is zero so plugging the numbers in, y = y o + 1 2 a y t 2 = 2.00 " 1 2 (9.80)(0.438) 2 # y = 1.06 m . 2. An 85.0kg base runner tries to steal second base. When he is 3.00m from the base he begins his slide at a speed of 9.00m/s. He comes to rest just as he touches the base. Find (a)his acceleration, (b)the average frictional force exerted by the ground on the sliding runner, and (c)the coefficient of friction between the runner and the ground. (a)Use the kinematic equation without time, v 2 = v o 2 + 2 a ( x " x o ) # = v o 2 " 2 ax o # a = v o 2 2 x o = ( " 9.00) 2 2(3.00) # a = 13.5 m / s 2 . (b)Applying the Second Law along the x-direction, " F x = ma x # F fr = ma = (85.0)(13.5) # F fr = 1150 N . (c)Applying the Second Law along the y-direction, " F y = ma y # F n \$ F g = # F n = F g = mg . Using the definition of COKF, μ k " F fr F n = ma mg = a g = 13.5 9.80 # μ k = 1.38 . x y x o = 0 y o = 2.00m x = 18.4m y = ? v ox = 42.0m/s v oy = 0 v x = 42.0m/s v y = ? a x = 0 a y = -9.80m/s 2 t = ? t = ? x y F fr F g F n given: x o = 3.00m x = 0 v o = -9.00m/s v = 0 a = ?...
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

S07s(3) - Physics 204A FINAL EXAM Chapters 1 14 Spring 2007...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online