# Math Help - velocity, acceleration vectors

1. ## velocity, acceleration vectors

I have two questions.

"As a figure skater spins, she raises and lowers her fingertip. This moving point has the parametrization
r(t)=<x,y,z>=<2Cos(2t), 2Sin(2t), 3 + 2t - t^2>
(The units are in feet and seconds.)
If a ring slips off her finger at time t=0 and falls due to gravity, write a parametric equation for the motion of this falling projectile."

I found the velocity vector to be v(t)=<-4sin(2t), 4cos(2t), 2-2t>. The position vector was given...and then I kinda got stuck.

And for the second question:

"What launch speed C is needed in order to kick a football over a goalpost that is 10 feet high and 100 feet away? Assume the launch angle is alpha=45 degrees. Use g=32 ft/sec^2. Solve for C in units of ft/sec.
Hint. The projectile motion is described by
x= Ct Cos(alpha)
and
y= Ct Sin(alpha) - 1/2g*t^2."

I drew a triangle and used the pythagorean theorum to find the hypotenuse to be 100.5, then found the value of sin(45) to be .707...and plugged in 100.5*.707 (aka, 71.1) into the x equation for C...and I can't really follow what I did after that. I ended up getting 71.1 ft/sec, which was wrong.

-bree

2. Originally Posted by bree56
I have two questions.

"As a figure skater spins, she raises and lowers her fingertip. This moving point has the parametrization
r(t)=<x,y,z>=<2Cos(2t), 2Sin(2t), 3 + 2t - t^2>
(The units are in feet and seconds.)
If a ring slips off her finger at time t=0 and falls due to gravity, write a parametric equation for the motion of this falling projectile."

I found the velocity vector to be v(t)=<-4sin(2t), 4cos(2t), 2-2t>. The position vector was given...and then I kinda got stuck.

And for the second question:

"What launch speed C is needed in order to kick a football over a goalpost that is 10 feet high and 100 feet away? Assume the launch angle is alpha=45 degrees. Use g=32 ft/sec^2. Solve for C in units of ft/sec.
Hint. The projectile motion is described by
x= Ct Cos(alpha)
and
y= Ct Sin(alpha) - 1/2g*t^2."

I drew a triangle and used the pythagorean theorum to find the hypotenuse to be 100.5, then found the value of sin(45) to be .707...and plugged in 100.5*.707 (aka, 71.1) into the x equation for C...and I can't really follow what I did after that. I ended up getting 71.1 ft/sec, which was wrong.

-bree
$r(t)= \left= \left<2\cos(2t), 2\sin(2t), 3 + 2t - t^2\right>$

$v(t)= \left<-4\sin(2t), 4\cos(2t), 2-2t\right>$

as the ring leaves, it experiences a acceleration g only in the vertical direction , assumed to be z .

$r(0) = \left<2,0,3\right>$

$v(0) = \left<0,4,2\right>$

for the ring, $t \ge 0$ ...

$v(t) = \left<0,4,2-gt\right>$

$r(t) = \left< 2, 4t , 3 + 2t - \frac{1}{2}gt^2 \right>$