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## Homework Statement

A weather rocket is launched straight upwards. The motor provides a constant acceleration for

18s, then the motor stops. The rocket altitude after 26s after launch is 5800m. What was the rocket's acceleration during the first 18 seconds?

## Homework Equations

[ tex ]x = x_0 + v_0 t + (1/2) a t^2

[/tex]

[ tex]v = v_0 + a t

[/tex]

## The Attempt at a Solution

This answer doesn't seem right to me. I wasn't confident working it out.

5800=.5(a)(18)^2-(.5)(9.81)(8)^2

162a=5800+313.92

162a=6113.92

a=37.7