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1. Oct 4, 2008

### uno

Hello,

I have been working on the following problems for hours and can't figure out where to begin. Any advice would be greatly appreciated.

-uno

1. An arrow is launched vertically upward from a crossbow at 97.9 m/s. Ignoring air friction, what is its instantaneous speed at the end of 10.0 s of flight?

(I figured out the instantaneous speed at -.2 m/s) A second part of the question asked for magnitude of instantaneous acceleration.

2. A raw egg is thrown horizontally straight out of the open window of a fraternity house. If its initial speed is 18 m/s and it hits ground 2.1 s later, at what height was it launched? At such low speeds air friction is negligible. [Hint: Even though it's been thrown horizontally, the egg falls vertically for 2.1 s exactly as if it were simply dropped.]

3. While rolling marbles on a horizontal window sill a youngster accidentally shoots one at 2.6 m/s out the open window. He sees it land in a flower pot on a neighbor's firescape 2.8 s later. How far beneath the sill is the pot?

4. Suppose you point a rifle horizontally directly at the center of a paper target 97 m away from you. If the muzzle speed of the bullet is 1020 m/s, where will it strike the target? Assume aerodynamic effects are negligible.

Thanks.

2. Oct 4, 2008

### Redbelly98

Staff Emeritus
uno, welcome to PF.

The kinematic equations for constant acceleration will help you here. Refer to those, and please include them in your posts ... as well as showing your attempt at solving the problem.

#1 is correct.

3. Oct 4, 2008

### uno

Hello,

For #1, the formula for constant acceleration is Vf-Vi/t. I know that Vi = 97.9, how do I find Vf? Thanks.

4. Oct 4, 2008

### Redbelly98

Staff Emeritus
You know what the acceleration and t are, so just solve for Vf.

But you already said Vf is -0.2 m/s in your first post.

5. Oct 4, 2008

### uno

Kinematics: Acceleration

I am having trouble with the following problem:

Suppose you point a rifle horizontally directly at the center of a paper target 97 m away from you. If the muzzle speed of the bullet is 1020 m/s, where will it strike the target? Assume aerodynamic effects are negligible.

I used the formula Tt = Square root of 2(distance)/9.81) to get a time of 4.45 seconds.

I am having some difficulty deciding what I am solving for Sx or Sy.

Any help would be appreciated.