# Thread: Far removed from Algebra and occasionally in need of guidance

1. ## Far removed from Algebra and occasionally in need of guidance

I trust that I will be able to join with those that are still very much acquainted with Algebra and receive answers that I have stored somewhere in my memory but can't find them.

2. ## Re: Far removed from Algebra and occasionally in need of guidance

If I am 6' tall and shoot a 3 point shot from 20', with a high arch, and it hits the rim and bounces off - how is the thrust determined? Maybe thrust isn't even the right word.

3. ## Re: Far removed from Algebra and occasionally in need of guidance

Originally Posted by Guinn
If I am 6' tall and shoot a 3 point shot from 20', with a high arch, and it hits the rim and bounces off - how is the thrust determined? Maybe thrust isn't even the right word.
I believe initial velocity (speed and angle relative to the horizontal) is what you mean by your use of the word "thrust".

The equations for projectile motion (ignoring air resistance which is negligible in this case) ...

$\Delta x = v_o \cos{\theta} \cdot t$

$\Delta y = v_o \sin{\theta} \cdot t - \frac{1}{2}gt^2$

where $\Delta x = 20 \, ft$

$\Delta y = 10 \, ft \, - \, height \, of \, ball \, release$

$t$ = time of ball travel in seconds

$g$ = magnitude of acceleration due to gravity ... about $32 \, ft/sec^2$

$v_o$ = ball's speed when released

$\theta$ = launch angle relative to the horizontal

note that there can be different solutions for the initial velocity's speed and launch angle, depending on the ball's time of flight.

recommend you "google" the topic projectile motion if you want more concrete examples

4. ## Re: Far removed from Algebra and occasionally in need of guidance

[QUOTE=skeeter;726699]I believe initial velocity (speed and angle relative to the horizontal) is what you mean by your use of the word "thrust".

The equations for projectile motion (ignoring air resistance which is negligible in this case) ...

$\Delta x = v_o \cos{\theta} \cdot t$

$\Delta y = v_o \sin{\theta} \cdot t - \frac{1}{2}gt^2$

where $\Delta x = 20 \, ft$

$\Delta y = 10 \, ft \, - \, height \, of \, ball \, release$

$t$ = time of ball travel in seconds

$g$ = magnitude of acceleration due to gravity ... about $32 \, ft/sec^2$

$v_o$ = ball's speed when released

$\theta$ = launch angle relative to the horizontal

note that there can be different solutions for the initial velocity's speed and launch angle, depending on the ball's time of flight.

recommend you "google" the topic projectile motion if you want more concrete examples[Does this have the formula for missing the shot and the velocity speed coming off the rim?]

5. ## Re: Far removed from Algebra and occasionally in need of guidance

Originally Posted by Guinn
[Does this have the formula for missing the shot and the velocity speed coming off the rim?]
you could calculate the velocity that the ball hits the rim, but to answer your question, no ... that would involve a physics concept known as "impulse" ... a bit more complex since the ball's angle of deflection would depend on the angle of incidence with the rim and would also depend on the quantity of the ball's energy absorbed by the rim.

in other words, simple collisions between objects in the real world really aren't so simple.