No Calculus allowed. I need some help. I'm really stuck...

Thank You.

http://i137.photobucket.com/albums/q...3_edited-1.jpg

http://i137.photobucket.com/albums/q...075_edited.jpg

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- January 31st 2009, 01:01 PMqbkr21Physics: Motion problem
No Calculus allowed. I need some help. I'm really stuck...

Thank You.

http://i137.photobucket.com/albums/q...3_edited-1.jpg

http://i137.photobucket.com/albums/q...075_edited.jpg - January 31st 2009, 01:31 PMskeeter
no calculus, huh?

since

max value for occurs when

min ball speed will occur when it is at the peak of its trajectory ... when ... min speed would therefore be ... ?

max height occurs at - January 31st 2009, 01:55 PMqbkr21
- January 31st 2009, 02:47 PMskeeter
**How did you figure all of these side details out like delta y = 0?**

if the projectile starts on the ground and ends on the ground, what is ?

**How do you determine which parts of the problem belong to X and which ones belong to y ?**

I don't understand the question ... what parts do you mean?

**For the last part can you at least show me the equations that I would use?**

I gave you the equation for time to the top of the projectile's trajectory ... you need to calculate for that time.

you could also ignore the time element and use this equation to calculate ...

at the top of its trajectory, - January 31st 2009, 03:49 PMqbkr21RE:
Question:

b) What is the minimum speed of the ball during this hole-in-one shot?

The book says the answer is 21.2 m/s

I set up my equation as follows...

(V_y)^2=(V_xy)^2-2g*delta(y)

0=V_xy)^2-1799.28

x^2=1799.28

x=42.4

***Where did I go wrong? - January 31st 2009, 06:49 PMskeeter
at the top of its trajectory, the projectile's y-component of velocity is zero ... so, it only has an x-component of velocity which is constant since there is no acceleration in the x-direction.