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March 30, 2015

March 30, 2015

Posted by **jo** on Saturday, September 18, 2010 at 6:45pm.

1.) A model rocket blasts off from the ground rising straight upward with a constant acceleration that has a magnitude of 86 m/s^2 for 1.70 seconds, at which point fuel runs out. Air resistance neglected What maximum altitude will the rocket reach?

2.)In reaching her destination, a backpacker walks with an average velocity of 1.34 m/s, due west. This average velocity of 2.68 m/s, due west,,turns around and hikes with an average velocity of 0.447 m/s due east. how far did she walk?

- physics -
**bobpursley**, Saturday, September 18, 2010 at 7:14pm1. Using the acceleration and time, calculate the velocity and height at fuel burnout. Then figure the height the rocket can go above this with that initial velocity. Add the heights to get the maximum altitude.

Here is a second way.

Energy put into rocket = change in PE

Force*distance=mgh

Mass*acceleration*distance= mgh

but distance= avgvelocity*time

= (Vf/2)*time=a*t^2/2

Mass*acceleration*acceleration*t^2=mgh

h= a^2/2g * t^2 where t is the time the rocket was accelerating.

check both ways.

2.Makes no sense to me.

- physics -
**jo**, Saturday, September 18, 2010 at 7:24pmokay. so regarding number 1. I will use the first method. So I will calculate using this formula?

x = v0t + 1/2at^2,

in which case.. x = 86(1.70) + 1/2(-10)(1.70)^2? I get x, which would be the position.. how do figure out the height over initial velocity? do i plus 0 in for v0?

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