3.74 s

+36.7 m/s

-36.7 m/s

68.5 m

+19.1 or -19.1 m/s

1. Helen Wheels launches a rocket in the air for a total of 7.48 seconds.  What time did it spend going up?  What is its initial velocity?  What is its final velocity? How high did it go?  What is the rocket's velocity at an elevation of 50.0 m?  (Why are there two answers?)

t=7.48s               it takes half of the time to go up and half to come down so

                       7.48s*(1/2)=    3.74s

a=v/t   a=9.8  t= 3.74s    9.8 is positive in this case because the rocket is going up

                              9.8*3.74s=  36.7m/s

coming down same equation but 9.8 is negative

                            -9.8*3.74=-36.7m/s

s=ut+1/2at^2       u=36.7m/s           t=3.74s    a= -9.8m/s^2

s=68.5m

v^2=u^2+2as   s= 50m  u= 36.7   a= -9.8

v=  19.1 and –19.1 m/s    because the rocket passes the spot on the way up and on the way down

525 s

14.2 km

2. A Honda driven by Cliff Jumper going 27 m/s is 2.1 kilometers (km) behind another Pinto driven by Ima Sloe going 23 m/s in the same direction.  What time will it take the Cliff to overtake Ima?  What distance will the Cliff travel in this time?

v1= 27m/s    v2=  23m/s   s=  2.1 km

v=s/t       v= 27m/s –23m/s  =4m/s                   Do not forget to convert

                                                        km to m     2.1*1000= 2100m

t=525s

v=27m/s   t= 525s  

v=s/t      s=14.2 km       

12.8 m/s

-24.3 m/s

8.3 m

-8.0 m/s

+5.1 m

3. Freda Dadark throws a ball up from the roof of a building at a height of 21.7 m.  It strikes the ground 3.78 seconds later.  What is the initial velocity of the ball?  With what velocity does the ball hit the ground?  How high above the building does the ball go?  What is the velocity and displacement 2.12 seconds after the ball was released?

u = ?

v = ?

a = -9.8

s = -21.7

t = 3.78

s = ut + 1/2at²

-21.7 = u(3.78) + 1/2(-9.8)(3.78)²   

-21.7 -  1/2(-9.8)(3.78)²   = u(3.78)

48.3136 = u(3.78), u = 12.7813 m/s

v = u + at, v = 12.7813 + (-9.8)*3.78 = -24.3 m/s

above the building

u = 12.7813, v = 0

v² = u² + 2as,  0² = 12.7813² + 2(-9.8)s , s = 8.33 m

Velocity and displacement at 2.12 s:

v = u + at, = 12.7813 + -9.8*2.12 = -8.0 m/s

s = ut + 1/2at² = 12.7813*2.12 + 1/2(-9.8)*2.12² = 5.1 m

4.74 s

-33.1m/s

110 m

4. Catona Hotinruff is ascending in a helicopter at a rate of 13.3 m/s.  At an elevation of 47.0 m, she drops a bagel out the window of the 'copter.  What time does the bagel take to reach the ground?  What is its velocity of impact?  How high is the helicopter when the bagel hits the ground?

V=13.3m/s    s=47m     a=-9.8        s=ut+1/2at^2

47=13.3t+1/2*-9.8t^2

t=4.74s

v=u+at            v=13.3+-9.8(4.47)

v=-33.15m/s

u=13.3m/s   t=4.74s

v=s/t   s=63.042   + the original 47m  =110m

49.2 m/s

808 m

5. Colin Host is driving his Ferrari 345 m behind my Tercel.  I am going 28.2 m/s, but the Colin overtakes me in 16.43 seconds.  How fast is the Ferrari going?  How far does the Ferrari travel before it overtakes me?

s=345m           v=28.2m/s   t=16.43s

v=s/t                      v1-28.2=345m/16.43s

v1=49.2m/s

v=s/t          49.2=s/16.43        s=808m

67.1 mph

6. Bob White is 12.5 miles from home.  If he drives the first 3.0 miles home at 35 mph, how fast does he need to drive the rest to average 55 mph

#6

To average 55 mph, the trip home should take  V=s/t so:  55mph=12.5/t 

Which means t=.2727272 hours.

The first 3 miles takes him v=s/t      35mph=3/t, t=.085714 hours

Now he has 12.5-3 = 9.5 miles to go and .27272-.085714=.141558 hours

So he needs to go v = s/t = 9.5/.141558 = 67.1 mph

5.94 s

60.7 m

-23.7 m/s

7. Justin Case fires an air rocket upward at 34.5 m/s from ground level.  Unfortunately (for him) the rocket lands on the top of a 32.0 m tall light tower on its way down.  What time is the rocket in the air?  How high  does the rocket go?  With what velocity does it strike the tower?

u= =34.5 m/s

a= -9.8 m/s

s= +32 m

use v² = u² +2as  so v² = (34.5)+2(-9.8)(32)

v= 23.73 m/s

Since the rocket hits the tower on the way down then v must be negative.  So now we know:     u==34.5 m/s, v=-23.7 m/s, a=-9.8 m/s, and s==32m

To find time in the air use v=u+at

-23.73=34.5 + (-9.8)t, t=5.94s

for how high we know

u=34.5 m/s, a=-9.8 m/s/s, and v=0 (top)

use v² = u² +2as 

 0= 34.5²+2(-9.8)s

s=60.7m

7.17 s

1.52 s (going up)

5.66 s (going down)

8. Molly Fayad pops a softball up that ends up being caught by the catcher at the same elevation as she hit it.  A spectator, using a sextant, and a range finder, determines that the ball went 63.0 m into the air at its highest point.  For what total time was the ball in the air?  At what times is the ball at an elevation of 42.0 m?

We know on the trip up:    u=?, v=0, s=63, a=-9.8 m/s/s

Use v² = u² +2as 

O= us+2(-9.8)(63)

U= 35.139m/s    positive because it starts out going up

To find time in air use twice the time to the top

U=35.139m/s  v=0  a= -9.8

v=u+at     0= 35.139 + -9.8t

t=3.5856s   twice that is 7.17s

s=42m   and the above velovity            v^2=u^2=2as    v^2=35.139^2+2(-9.8)(42)

v=20.288m/s   positive on the way up, negative on the way down                                     

going up  

u=35.139m/s      v=20.288m/s    a=-9.8      v=u+at      20.288=  35.139+-9.8t

t=1.52s

going down

v=u+at           -20.288=35.139+-9.8t      t=5.66s

6.65 m/s

(6.52 is wrong)

9. Austin Tascious sends a bowling ball down a 16.5 m long lane, and we hear the sound of impact 2.53 seconds later.  If the speed of sound is 343 m/s, what speed did the bowling ball move?

T=2.53s         Vsound=343m/s

Sound takes     v=s/t          343=16.5/t       =.0481s

2.53-.0481=2.482s    for the ball to reach the pins

the ball must be going      v=s/t     16.5/2.482  =6.65m/s

57.5 m

10. Ali Katz is taking a dripping paint bucket back to the store, so he holds the paint bucket out the car's window, letting the paint drip onto the ground.  The bucket is dripping at some unknown rate, and his car accelerates at some unknown rate, starting from rest.  If the first drop drips when his car's velocity is zero, the second drop is 2.3 m down the road, how far is the sixth drop down the road?

Second drop       t2=?        A=?       s2=2.3

We also know the fifth drop      t5=4t2    (it dripped four times later than the second drop    a=?     s=?

S=ut+1/2at^2    since u=0   s=1/2at^2                     A=B    &    C=D

S2=1/2at2^2   and s5=1/2a(4t2)^2                  then         A/C=B/D

Both “a” s are the same so we can  solve

2.3m=1/2a(t2)^2             =       (t2)^2            =    1     

s5     =1/2a(4t2)^2                     16(t2)^2            16

2.3/s5  =   1/16      s5= 2.3*16       =  36.8m