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*T2

2

5) Vf2 = Vi2 + 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*AT2

0=8T+ ½*-9.8 T2

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

= 82+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*AT2

D=0*1.5+ ½*-9.8 1.52

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)

Vf2= 42+2(-9.8)(-4)

Vf2=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*T2

.32 = ½*9.8*T2

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*AT2

0=Vi*3+ ½*-9.8 32

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*AT2

D=0T+ ½*-9.8*1.52

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

02=14.72 +2(-9.8)D

02= 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/s2 .

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*AT2

-24000=0T+ ½*-3.7*T2

-24000= -1.85 T2

T2 = 24000/1.85 so T= 113 seconds or 1.89 minutes

Vf2=02 +2(-3.7)(-24000)

Vf2=177600

Vf=421.4 m/s

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