A test rocket is fired straight up from rest with a net acceleration of 20.0 m/s2.After 4.00 seconds the motor turns off, but the rocket continues to coast upward.What maximum elevation does the rocket reach?Neglect air friction and express your answer in meters.

To solve this problem, we need to break it down into two parts: the ascent while the motor is on and the ascent while the rocket is coasting upward.

1. Ascent while the motor is on:

We can use the equation of motion: d = ut + 0.5at^2 Where: d = distance u = initial velocity = 0 (since the rocket starts from rest) a = acceleration = 20 m/s^2 t = time = 4 seconds

Substituting these values into the equation, we get: d = 04 + 0.5204^2 = 0 + 0.520*16 = 160 meters

2. Ascent while the rocket is coasting upward:

When the motor turns off, the rocket will continue to move upward until its velocity becomes zero. We can find the time it takes for this to happen using the equation: v = u + at Where: v = final velocity = 0 (since the rocket eventually stops) u = initial velocity = accelerationtime = 204 = 80 m/s (from the first part of the journey) a = acceleration = -9.8 m/s^2 (since gravity is now acting against the rocket)

Solving for t, we get: 0 = 80 - 9.8*t t = 80 / 9.8 = 8.16 seconds

Now we can find the distance the rocket travels during this time using the equation of motion again: d = ut + 0.5at^2 d = 808.16 + 0.5(-9.8)*8.16^2 = 326.4 meters

Adding the distances from the two parts of the journey, we get the maximum elevation the rocket reaches: 160 meters + 326.4 meters = 486.4 meters

So, the rocket reaches a maximum elevation of 486.4 meters.