__NAME____________

__Gravity and constant acceleration__

__ __

If an object is thrown up and falls down then:

Time up = Time down = half entire time.

Dis at bottom is 0.

V at top = 0

Vi up = -Vf down

1) D = (__Vi+Vf)__ * T 3)
Vf = Vi + A*T

2

2) Vavg = (__Vi+Vf)__ 4)
D = Vi*T + ½*A*T^{2}

2

5)
Vf^{2 }= Vi^{2} + 2*A*D

G Problems:

1) If I throw up a baseball straight into the air at 8 m/s, (assume g=-9.8 )

a)what is its velocity at the top of its journey?

*Vf = 0 at top*

b) How much time does it take to go all the way up?

*Vf=0, Vi=8, A = -9.8*

*Vf=Vi+AT 0=8-9.8T 9.8T=8 T=8/9.8=.81632
seconds*

c) How much time does it take to go down?

*Same amount of time T= .81632
seconds*

*Total time up and down, D=0,
Vi=8, A=-9.8*

*D=ViT + ½*AT ^{2} *

*0=8T+ ½*-9.8 T ^{2}*

*0=8-4.9T T=8/4.9
=1.632 seconds for whole trip, ½ = .81632 seconds*

d) What is its final velocity when it hits the ground?

*Vf=Vi+AT*

*Vf=8+(-9.8)(1.632) = 8-16 =
-8 m/s*

* *

*Or Vf ^{2}=Vi^{2}
+2AD*

* =
8 ^{2}+2(-9.8)(0)*

*so Vf=+/- 8 m/s*

2) A pebble is dropped down a well and hits the ground 1.5 seconds later. What is the displacement from the edge of the well to the waters surface?

* Vi=0
(dropped)*

*A= -9.8*

*T=1.5*

*D=???*

*D=ViT + ½*AT ^{2} *

*D=0*1.5+ ½*-9.8 1.5 ^{2}*

*D= -4.9*2.25 = -11.025 m (downwards)*

3) A gymnast is practicing a dismount from the high bar that is 4 meters off the ground, and swings up with a velocity of + 4 m/s . How fast will she be going when she hits the ground?

*Vi=4, A = -9.8 , D= -4 m (because up is positive, and
she’s ending up below the start)*

* *

*Vf ^{2}= 4^{2}+2(-9.8)(-4)*

*Vf ^{2}=16+78.4=94.4*

*so Vf=+/- 9.7 m/s or -9.7 m/s
downwards*

4) I drop a meterstick and catch it at the 32 cm mark. What is my reaction time?

*Vi=0, A = - 9.8 , D = -32 cm = - .32 m, T= ???*

*D= ½*g*T ^{2}*

*.32 = ½*9.8*T ^{2}*

*T= SQRT ( 2*.32/9.8 ) = SQRT (.0653) = .256 seconds*

5) I jump straight up and hit the ground 3 seconds later.

How fast was I going when I started?

*Vi=????, T=3, D=0 (hit ground), A = -9.8*

*D=ViT + ½*AT ^{2} *

*0=Vi*3+ ½*-9.8 3 ^{2}*

*0=3Vi -4.9 * 9*

*44.1 = 3Vi Vi=14.7
m/s, Vf = -14.7 m/s*

* *

*or at top of journey, Vf=0,
T=1/2 total t= 1.5, A = -9.8*

* *

*Vf=Vi + A T 0=Vi -9.8*1.5 Vi= 14.7 m/s*

* *

What is the total DISTANCE I traveled (not displacement)

*Total distance equals distance up plus distance down,
or twice distance down.*

*Starting at top T =1.5, Vi=0, A = -9.8, D=??*

*D=ViT + ½*AT ^{2} *

* D=0T+
½*-9.8*1.5 ^{2}*

* D=11.025m
so total distance = 22.05 m*

*Or use #5 for first half of journey, Vi=14.7, Vf=0, A=
-9.8, T=1.5*

* 0 ^{2}=14.7^{2} +2(-9.8)D*

*0 ^{2}= 216.09+ -
19.6D*

*216.09=19.6D D=11.025m
up, so total D is 2*11.025 = 22.05 m*

6) A robot probe drops a camera off the rim of a 24 km deep
crater on Mars, where the free fall acceration is -3.7 m/s^{2} .

Find the time required for the camera to reach the crater floor and the velocity with which it hits.

*D= -24 km = -24000 m, Vi=0, A = - 3.7 T =???*

*D=ViT + ½*AT ^{2} *

* -24000=0T+
½*-3.7*T ^{2}*

*-24000= -1.85 T ^{2} *

* T ^{2} = 24000/1.85 so T= 113 seconds or 1.89 minutes*

* *

*Vf ^{2}=0^{2}
+2(-3.7)(-24000)*

*Vf ^{2}=177600*

*Vf=421.4 m/s*

Sketch the distance time and velocity time graph as compared
to Earth.