Can I have help with part (c)? How would I do it? (Worried) Thanks in advance.

EDIT:Can I have help with part (d) too. Thanks.

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- Apr 15th 2008, 08:11 AMSimplicityKinematics
Can I have help with part (c)? How would I do it? (Worried) Thanks in advance.

**EDIT:**Can I have help with part (d) too. Thanks. - Apr 15th 2008, 09:11 AMMagnusc
C:

Horizontal M*V doesn't change:

Initial horisontal M*v = 100 * cos(60) * 80 = 4000

M*v after crasch = 60 * 80 - 40 * x = 4000 => x = 20

kind regards - Apr 15th 2008, 10:14 AMtopsquark
There are two SUVAT formulas here, one in the x direction and one in the y:

$\displaystyle x = v_{0x}t = 80t~cos(60)$

and

$\displaystyle y = 20 + 80t~sin(60) - 4.9t^2$

We can find the time for max height by noting that the y component of the velocity of the rocket

$\displaystyle v_y = 80~sin(60) - 9.8t$

is equal to 0 at that time.

Plugging that time into the y equation gives the max height.

For the explosion stage of the problem, note that the explosion is internal, meaning that the momentum of the rocket and its pieces is conserved. Thus:

$\displaystyle 100v_0 = (60~kg)(80~m/s) + (40~kg)v$

($\displaystyle v_0$ is the velocity at max height, which will simply be the x component of the velocity.)

Solve for v.

For the last question, you know v and you know the height. So you can set up your x and y equations again and find out where Q lands. (You will need both the x and y coordinates of the max height to start you off on this.)

-Dan