Answer:
a) 200 m
b) 100 m/s
c) 709 m
d) -118.2 m/s
e) 26.24 s
Explanation:
The rocket flies upward with constant acceleretion.
The equation for position under constant acceleration is:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
Y0 = 0
V0 = 0
Y(t) = 1/2 * 25 * t^2
Y(t) = 12.5 * t^2
And speed under constant acceleration:
Vy(t) = Vy0 + a * t
Vy(t) = 25 * t
It burns for 4 s and runs out of fuel
Y(4) = 12.5 * 4^2 = 200 m
V(4) = 25 * 4 = 100 m/s
Form t = 4 the rocket will coast, it will be in free fall, affected only by gravity
It will be under constant acceleration. These new equations will have different starting constants.
Y(t) = Y4 + Vy4 * (t - 4) + 1/2 * g * (t - 4)^2
Vy(t) = Vy4 + g * (t - 4)
When it reaches its highest point it will have a speed of zero.
0 = Vy4 + g * (t - 4)
0 = 100 - 9.81 * (t - 4)
100 = 9.81 * (t - 4)
t - 4 = 100 / 9.81
t = 10.2 + 4 = 14.2 s
At that moment it will have a height of:
Y(14.2) = 200 + 100 * (14.2 - 4) - 1/2 * 9.81 * (14.2 - 4)^2 = 709 m
The rocket will fall and hit the ground:
Y(t) = 0 = 200 + 100 * (t - 4) - 1/2 * 9.81 * (t - 4)^2
0 = 200 + 100 * t - 400 - 4.9 * (t^2 - 8 * t +16)
0 = -4.9 * t^2 + 139.2 * t -278.4
Solving this equation electronically:
t = 26.24 s
At that time the speed will be:
Vy(t) = 100 - 9.81 * (26.24 - 4) = -118.2 m/s
