**A model rocket is fired vertically from the ground with a constant acceleration of 45.3 m/s^2 for 1.46 s at which time its fuel is exhausted. What is the maximum height (in m) reached by the rocket? The answer is 271 m but can someone show me the work please? thanks!**

**Answer:**

your acceleration a=45.3 m/s^2 and is constant that means it is due to the fuel and gravitational field is considered in it so you don’t have to worry about gravity for case 1.

case 1: (from t=0 s to t=1.46 s)

initial velocity =0 m/s. time t=1.46 s

let s1 be your distance covered with constant acceleration a.

by using s=u*t +(0.5) a(t^2)
s1=48.28 m.
now final velocity v1 is given by: v=u+a*t

so v1=66.138 m/s.

case 2:Now as the fuel is over there won’t be any acceleration but there will be deceleration due to gravity but rocket will move upwards as it has a initial velocity u2 = v1.At highest point velocity will be 0m/s thus our final velocity v2 =0m/s

thus the distance covered upwards can be given by s2 :

using v^2=u^2 +2

*a*s

now a=-g=-9.8 m/(s^2)

we get :

0 = (66.138)^2 - 2*(9.8)*s2

s2 = 223.175 m.

so adding s1 and s2

total distance covered =271.45 m.