# The Momentum Principle

• Sep 24th 2009, 01:59 PM
TGS
The Momentum Principle
I am completely lost... I don't really understand the momentum principle that well so I need someone to explain this to me.

"A ball is kicked from a location < 8, 0, -7 > (on the ground) with initial velocity < -9, 16, -4 > m/s. The ball's speed is low enough that air resistance is negligible.
What is the velocity of the ball 0.2 seconds after being kicked?"

And this as well:

"Now consider a different time interval: the interval between the initial kick and the moment when the ball reaches its highest point. We want to find how long it takes for the ball to reach this point, and how high the ball goes.
What is the y-component of the ball's velocity at the instant when the ball reaches its highest point (the end of this time interval)?"
• Sep 24th 2009, 04:28 PM
skeeter
Quote:

Originally Posted by TGS
I am completely lost... I don't really understand the momentum principle that well so I need someone to explain this to me.

"A ball is kicked from a location < 8, 0, -7 > (on the ground) with initial velocity < -9, 16, -4 > m/s. The ball's speed is low enough that air resistance is negligible.
What is the velocity of the ball 0.2 seconds after being kicked?"

And this as well:

"Now consider a different time interval: the interval between the initial kick and the moment when the ball reaches its highest point. We want to find how long it takes for the ball to reach this point, and how high the ball goes.
What is the y-component of the ball's velocity at the instant when the ball reaches its highest point (the end of this time interval)?"

assuming that I have interpreted your given coordinates correctly ...

in the vertical direction, the acceleration due to gravity is \$\displaystyle -9.8 \, m/s^2\$

\$\displaystyle v(t) = <-9 \, , \, 16-9.8t \, , \, -4>
\$

at the top of its trajectory, \$\displaystyle v_y = 0\$

position, \$\displaystyle r(t) = <8-9t \, , \, 16t-4.9t^2 \, , \, -7-4t>\$