ANSWERS
Quiz 2 - Chapter 2 
Ch. 2:   33, 34, 36, 37, 41, 42, 44, 57
Click here for the  correction to the hint to #68, Quiz 1. Scroll down to #68.
33. Use change in y = V1yt - (1/2)gt2, where the initial velocity is zero. Note that we are choosing UP to be positive, since the acceleration shown is negative.  
34. Vy = V1y - gt, where the initial velocity is zero. Note that we are choosing UP to be positive, since the acceleration shown is negative.  
 36.  

(a) Vy = V1y - gt, where the initial velocity is zero. Note that we are choosing UP to be positive, since the acceleration shown is negative.  To find it takes to reach the top, set the final velocity to zero and solve for t. Then plug into 


change in y = V1yt - (1/2)gt2.  Note the initial velocity is given.

ALTERNATIVE SOLUTION: 

Use Vy2 = V1y2 - 2g(change in y) 
Set the final velocity to zero and find change in y. 
(b) Short cut:  multiply the time you got in (a) by 2. 
Longer way:  Use change in y = V1yt - (1/2)gt2  , set left hand side of equation to zero, and solve for t like we did in class. 
 37. 
(a) Use (change in y) = V1yt - (1/2)gt2        and set the left hand side of the equation to zero. Then solve for V1y  after substituting the time t given. 
(b) To find the maximum height, see 36 (a), or realize that the time to reach the top is 3.0/2 = 1.5 seconds. Plug that time into:  (change in y) = V1yt - (1/2)gt2 
 41.  Use (change in y) = V0t - (1/2)gt2  , set left hand side of equation to zero, and solve for t like we did in class. 
Finally, use Vy = V0 - gt, and plug in the t you got in the first part of this hint. When you do, you will find  Vy = -V0. 
 42. 
(a) Use: Vy2 = V1y2 - 2g(change in y) .  Plug in the change in y given, and find the speed; take the positive square root. Note that the initial velocity is given. Note that the negative square root corresponds to when the object is coming back down at the same height ! 
(b) and (c):  (change in y) = V1yt - (1/2)gt2    after plugging in the value of the change in y given. You will get two solutions to the quadratic equation  corresponding to rising and falling at the same height. See comment part (a). 
 44.  IN THE EQUATIONS BELOW, WE ASSUME THAT UP IS THE POSITIVE DIRECTION; HENCE DOWN IS NEGATIVE, WHICH EXPLAINS CERTAIN NEGATIVE SIGNS. 

IF YOU ASSUME DOWN IS POSITIVE, THEN YOU'D  RE-WRITE THE EQUATIONS WITHOUT THOSE NEGATIVE SIGNS. 

Method 1:

Just use two equations:
(A)  V1y2 = -2g(y1 - y0)   and (B)  -2.2 = V1y (0.28 s) - (1/2)g(0.28 s)2.

Solve (B) for V1y .  Then substitute into (A) to find y1 - y0. 

Method 2 (more difficult):

Note:

(I) -2.2 (m) = [(V1y + V2y)/2](0.28 s) . V1y  = velocity at the window top and  V2y = velocity at window  bottom. 
We want  

(II)  y1 - y0 = -V1y2/2g . You need  
Note you can use: 

(III)V2y2 = V1y2 - 2g(-2.2 m) . 

Solve (I) and (III) simultaneously for  V1y and  V2y  then substitute into (II). 
 more hints later