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!
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=ut +(0.5)a(t^2)
now final velocity v1 is given by: v=u+at
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 +2as
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.