Question

A rocket is fired straight upward, starting from rest with an acceleration of 25.0 m/s^2. It runs out of fuel at the end of 4.00 s and continues to coast 25.0 m/s^2. It runs out of fuel at the end of 4.00 s and continues to coast upward, reaching a maximum height before falling back to Earth. (a) Find the rocket’s height when it runs out of fuel; (b) find the rocket’s velocity when it runs out of fuel; (c) find the maximum height the rocket reaches; (d) find the rocket’s velocity the instant before the rocket crashes into the ground; and (e) find the total elapsed time from launch to ground impact.

Answer 1

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