# Need help with homework due on the december 6.

1. Dec 6, 2004

### Beowulf

I'm unsure of what equations to use for this certain questions. I'd very much apreciated if anyone could tell me which equation is to be used for each individual problem. This is also a study review for an exam that's coming up very soon.

1. A bullet traveling 8.0 x 10^2 10^2 m/s horizontall hits a target 180 m away. How far does the bullet fall before it hits the target?

2. A student threw a ball herizontally out of a window 8.0 m above the ground. It was caught by another student who was 10.0 m away. What was the initial velocity of the ball?

3. A baseball was hit at 45 m/s at an angel of 45(degrees) above the horizontal.
a)How long did it remain in the air?
b)How far did it travel horizontally?

4. A camper dives from the edge of a swimming pol at water level with a speed of 8.0 m/s at an angle of 30.0(degrees above the horizontal.
a) How long is the diver in the air?

b) How high does the diver go?

c) How far out in the pool does the diver land?

5. An amusement park ride consists of turntable of 2.0 m radius turning at .70 rev/s about a vertical axis. If a 70.0 kg child sits at the outer edge of the turntable, what force is necessary to keep the child from sliding off?

6.A 90.0-kg person is spinning around on the equator of planet X, which is rotating at 3.2 x 10^3 km/h. The centripetal force holding the person in place is 6.0 N. What is the radius of planet X?

7. A satellite orbits a planet at 4.0 x 10^3 m/s, and the acceleration of gravity on the satellite is .58 m/s^2. What is the diameter of the orbit?

8. What is the period of a pendulum 3.0 m long?

9. On earth the length of a pendulum with a period is 1.0 s is .25 m. What is the length of the pendulum with a period of 1.0 s on the moon, where the acceleration of gravity is 1/6 that of the earth?

Thank you.

2. Dec 6, 2004

### BobG

The first few break the velocity into two components: one horizontal, one vertical. That means you wind up using two equations. Both are essentially the same equation except the vertical component is affected by gravity, the horizontal component isn't. In other words, your vertical component has acceleration as part of the equation, acceleration is zero for the horizontal component.

Number 7 balances centripetal acceleration against centrifugal, but, if you know Newton's Universal Law of Gravitation, there's an easier way than the way they intend for you to solve it.

Last edited: Dec 6, 2004
3. Dec 6, 2004

### Beowulf

ok...but what exactly are the equations?

4. Dec 6, 2004

### Skotster

V=V0+a*t
d=V0*+(1/2)*a*t
V2=V02+2*a*d

V=Final velocity
V0=Initial Velocity
d=Distance
a=Acceleration
t=time

5. Dec 7, 2004

### Tom McCurdy

Keep in mind you treat the equations seperatly for X and Y componets

$$V_{y}=V_{oy}+a_{y}*t$$
$$V_{x}=V_{ox}+a_{x}*t$$
$$x=V_{ox}+\frac{1}{2}a_{x}*t^2$$--SQUARED!!!!!!!!!!!!!
$$y=Y_{oy}+\frac{1}{2}a_{y}*t^2$$--SQUARED!!!!!!!!!!!
$$V^2=Vo^2+2*a*x$$-break up in same fashion

6. Dec 7, 2004

### Tom McCurdy

Wrong!!!!

AGAIN Don't USED HIS DISTANCE EQUATION IT IS WRONG the t needss to be t^2
d=V0*+(1/2)*a*t^2=correct form
d=V0*+(1/2)*a*t=wrong form

7. Dec 7, 2004

### Tom McCurdy

Also to break up for componets you will need to take Vo*sin(angle) with respect to horizontal for y componet and Vo*cos(angle) with respect to horiztonal for x componet

gl... i am out for tonight i think

8. Dec 7, 2004

### Tom McCurdy

hehe one more thing before we go... tis odd that I meet beowulf as I am just starting to read the poem... kind of intersting.

9. Dec 7, 2004

### Skotster

sorry about the typo forgetting to add the "^2" after t in my equasions