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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.