1. Oct 28, 2009

### merceb49

A toy car runs off the edge of a table that is 1.325 m high. The car lands 0.425 m from the base of the table.
How long did it take for the car to fall off the table?
How fast was the car going when it fell off the table?

and

An arrow is shot at 30.0° above the horizontal. Its initial speed is 49 m/s and it hits the target.
What is the maximum height the arrow will attain?
The target is at the height from which the arrow was shot. How far away is it?

I was gone when my teacher went over this and my tutor did not understand this. PLEASE ANYONE I need suggestions hints anything to help me get these 4 questions correct. this is make or brake, please help soon. Please

2. Oct 28, 2009

### Delphi51

They are called projectile motion problems. Likely you can find an example in your text or on the web. Basically you
separate the initial velocity into horizontal and vertical parts. Then write two headings:
Horizontal and Vertical. Decide in each case whether you have constant speed or accelerated motion and write the appropriate formula(s). Put in the numbers or expressions for all known quantities and look for an equation you can solve.

3. Oct 28, 2009

### pr0blumz

If you have an initial velocity and an angle then you should be able to use you trig. to solve for this.

4. Oct 28, 2009

### merceb49

using trig gives me 24.5 and that is a velocity, not a measure of height.
I would have plugged into the appropriate formulas, the only problem is I do not know any necessary formulas. could i get some help with that?

5. Oct 28, 2009

### pr0blumz

v_o * sin theta = y
v_o * cos theta = x

6. Oct 28, 2009

### merceb49

with that i get .3 and .9. i dont understand how trig will lead me to the height when all i have to measure with is a velocity

7. Oct 29, 2009

### ApexOfDE

I think you dont understand the solutions posted above. Lets us consider again you problem. Arrow's initial speed is 49m/s with 300 from horizontal. From this, you have:

v0x = v0 * cos(theta) and v0y = v0 * sin(theta).

Horizontal speed is constant but vertical speed is not. Vertical speed will be decreased from highest value (v0y) to zero (when arrow reaches highest point in the air).

From textbook, you have: v(t)^2 - v0^2 = 2 * a * s(t) . In your problem, v(t) is vertical velocity of arrow at highest point s(t) ==> v(t) = 0. Acceleration is g (downward), meanwhile motion is upward ==> a = -g.

After you have s(t), you can find t ==> x = v0x * t

8. Oct 29, 2009

### merceb49

oh ok, and S is representing? physics is not my calling...

9. Oct 29, 2009

### pr0blumz

s=height

10. Oct 29, 2009

### merceb49

i thank you all for the help, i just dont understand this at all. thank you for the time