How High Does a Rocket Go if It Accelerates at 79.0 m/s² for 1.90 Seconds?

[harrism1]

Homework Statement

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 79.0 m/s2 for 1.90 seconds, at which point its fuel abruptly runs out. Air resistance has no effect on its flight. What maximum altitude (above the ground) will the rocket reach?

2. S=.5at2^2

The Attempt at a Solution

S= .5(79)(1.90)2
=142.595

S=142.595^2/(2*9.8)
=1037.42

142.595+1037.42=1080m... only 3 sig figs

It says its wrong?

 

[rl.bhat]

The second displacement is wrong. Find the velocity after 1.9 seconds. That will be the initial velocity for the second stage. Final velocity of the second stage is zero.

 

[Fightfish]

I don't understand the second equation. Clearly it is from $$v^{2} = u^{2} + 2as$$, giving $$s = u^{2} / 2g$$.
But why are you substituting the earlier displacement into u?

 

[pgardn]

Homework Statement

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 79.0 m/s2 for 1.90 seconds, at which point its fuel abruptly runs out. Air resistance has no effect on its flight. What maximum altitude (above the ground) will the rocket reach?

2. S=.5at2^2

The Attempt at a Solution

S= .5(79)(1.90)2
=142.595

S=142.595^2/(2*9.8)
=1037.42

142.595+1037.42=1080m... only 3 sig figs

It says its wrong?

You need to find the velocity of the rocket when the fuel gives out. Then you can use the free fall equation for the second part of the trip (only g accel). Max height V final = 0 m/s and remember that the initial velocity needs to be determined, which is the same as the final velocity when the rocket has that big acceleration up that gives out with the fuel.

And I see my post is old news...