Question

Motion along a curve. A ball starting at (0,0) passes through the point (5,2) after 2 seconds. Find Vo and the angle.

Answer 1

Having to do with projectiles.

So The equations I have to work with are

\[\large Height = \frac{(v_o sin(\alpha))^2}{2g}\]
\[\large Time = \frac{2v_o sin(\alpha)}{g}\]
\[\large Range = \frac{v_o^2 sin(2\alpha)}{g}\]

Answer 2

projectiles, so the equation will be a quadratic right?

Answer 3

I believe so yeah, But wouldn't it only be a quadratic if we dont worry about the parameter time?

Answer 4

no, the parameter is time is quadratic as a result of a constant acceleration

Answer 5

a(t) = -g

v(t) = -gt + vo

s(t) = -gt^2/2 + vo t + so

Answer 6

if we are at an angle, then the upward motion (since vertical is linear) has vo sin(a) as the portion of velocity in the upward direction

Answer 7

x = vo cos(a) t for horizontal ploting

Answer 8

Right..so the projectile path would be

\[\large x(t) = (v_o cos \alpha)t\]
\[\large y(t) = (v_o sin \alpha)t - \frac{1}{2}gt^2\] right?

Answer 9

yes