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!)
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
2 answers
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)
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)