Kinematics question - Rocket launch

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Kinematics question -- Rocket launch

Homework Statement


A rocket starts from rest and moves upward from the surface of the earth. For the first 10s of its motion, the vertical acceleration of the rocket is given by ay=(2.70m/s3)t, where the +y-direction is upward.
A.What is the height of the rocket above the surface of the earth at t = 10s?
B.What is the speed of the rocket when it is 200m above the surface of the earth?


Homework Equations


y=y(not)+V(not y)*t-(1/2)*g*t^2
v=V(not y)+a*t

The Attempt at a Solution



I am stuck on part A and need some help. I used the first equation to find an answer that was wrong. Here is my work:

ay=2.7*(10) =27m/s^3

I then divided that by 9.8 m/s^2 and then multiplied it by 100.

y=0+0-(1/2)*(27/9.8)*(10)^2
y=-860 m
I don't believe the actual answer is going to be negative if the up direction is considered (+). I tried putting (+) 860 in and it said it was wrong. If someone could point me in the right direction for Part A and B that would be much appreciated.
 

Answers and Replies

  • #2
gneill
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You're given the vertical acceleration of the rocket so it must already take into account all the forces acting, including that due to gravity. Since acceleration is varying over time (it's not constant) you must abandon all your kinematic equations and go back to basics. In other words, a little calculus is called for.
 

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