# Physics: Motion problem

• Jan 31st 2009, 01:01 PM
qbkr21
Physics: 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
• Jan 31st 2009, 01:31 PM
skeeter
no calculus, huh?

since $\displaystyle \Delta y = 0$

$\displaystyle 0 = v_{yo}t - \frac{1}{2}gt^2$

$\displaystyle 0 = t\left(v_{yo} - \frac{1}{2}gt\right)$

$\displaystyle v_{yo} = \frac{1}{2}gt$

$\displaystyle t = \frac{2v_{yo}}{g} = \frac{2v_o\sin{\theta}}{g}$

$\displaystyle \Delta x = v_x \cdot t$

$\displaystyle \Delta x = v_o\cos{\theta} \cdot \frac{2v_o\sin{\theta}}{g}$

$\displaystyle \Delta x = \frac{v_o^2 \cdot 2\sin{\theta}\cos{\theta}}{g}$

$\displaystyle \Delta x = \frac{v_o^2 \sin(2\theta)}{g}$

max value for $\displaystyle \Delta x$ occurs when $\displaystyle \sin(2\theta) = 1$

$\displaystyle \Delta x_{max} = \frac{v_o^2}{g}$

min ball speed will occur when it is at the peak of its trajectory ... when $\displaystyle v_y = 0$ ... min speed would therefore be ... ?

max height occurs at $\displaystyle t = \frac{v_{yo}}{g}$
• Jan 31st 2009, 01:55 PM
qbkr21
Quote:

Originally Posted by skeeter
no calculus, huh?

since $\displaystyle \Delta y = 0$

$\displaystyle 0 = v_{yo}t - \frac{1}{2}gt^2$

$\displaystyle 0 = t\left(v_{yo} - \frac{1}{2}gt\right)$

$\displaystyle v_{yo} = \frac{1}{2}gt$

$\displaystyle t = \frac{2v_{yo}}{g} = \frac{2v_o\sin{\theta}}{g}$

$\displaystyle \Delta x = v_x \cdot t$

$\displaystyle \Delta x = v_o\cos{\theta} \cdot \frac{2v_o\sin{\theta}}{g}$

$\displaystyle \Delta x = \frac{v_o^2 \cdot 2\sin{\theta}\cos{\theta}}{g}$

$\displaystyle \Delta x = \frac{v_o^2 \sin(2\theta)}{g}$

max value for $\displaystyle \Delta x$ occurs when $\displaystyle \sin(2\theta) = 1$

$\displaystyle \Delta x_{max} = \frac{v_o^2}{g}$

min ball speed will occur when it is at the peak of its trajectory ... when $\displaystyle v_y = 0$ ... min speed would therefore be ... ?

max height occurs at $\displaystyle t = \frac{v_{yo}}{g}$

How did you figure all of these side details out like delta y = 0?

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

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

Thanks,
qbkr21
• Jan 31st 2009, 02:47 PM
skeeter
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 $\displaystyle \Delta y$?

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 $\displaystyle \Delta y$ for that time.

$\displaystyle \Delta y = v_{yo}t - \frac{1}{2}gt^2$

you could also ignore the time element and use this equation to calculate $\displaystyle \Delta y$ ...

$\displaystyle v_{yf}^2 = v_{yo}^2 - 2g(\Delta y)$

at the top of its trajectory, $\displaystyle v_{yf} = 0$
• Jan 31st 2009, 03:49 PM
qbkr21
RE:
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?
• Jan 31st 2009, 06:49 PM
skeeter
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.

$\displaystyle v_x = v_o\cos{\theta}$