Question

Here's a physics problem. I don't care about the answer, i just want to know how to do it. don't have to be too detailed.. but i don't really kno how to start this

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?
What I did: okay. 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 plug 0 in for v0?
"BUMP FOR PURSLEY or anyone else" I got acceleration because its a constant.. its either 10, -10, etc. So I guess if its going up, acceleration should be 9.8? Its sorta given. I used 86 m/s as the velocity. as i said.. im having trouble setting it up

Answer 1

the initial velocity (v0) should be in the units of m/s! not m/s^2! at ground level, we could assume that the rocket is at rest (0 m/s)

as for acceleration, the units should be m/s^2 (given in the problem!)

Answer 2

after the 1.70 sec, the rocket stops accelerating. Find the final velocity. Then plug it back in the equation as the initial velocity. set vf=0 and a=-9.8m/s^2 (gravity) and solve for time.

vf = vi + at

plug those values back into the distance formula you provided and solved for the distance following the 1.70sec.

Add this distance to the other distance (0 sec to 1.70 sec)