Question

*brainlest* An object accelerates from rest with a constant acceleration of 2m/s^2. How far will it have moved after 9s? Equation: xx = x0 + v0t + 1/2 at^2

Answer 1

Answer: You throw an object upwards from the ground with a velocity of 20m/s and it is subject to a downward

acceleration of 9.8 m/s2. How high does it go?

Given Formula Set up Solution

vi=20 m/s

vf=0 m/s

a=-9.8 m/s2

x=?

vf

2 = vi

2 + 2a x (0 m/s)2= (20 m/s)2 + 2 (-

9.8 m/s2) x

x=20.4 meters

Explanation: i known/ i hoped that helped.