Math Problem Statement

A test rocket at ground level is fired straight up from rest with a net upward acceleration of 20 m/s2. After 4.0 s, the motor turns off but the rocket continues to coast upward with insignificant air resistance. What maximum elevation does the rocket reach?

Solution

To find the maximum elevation reached by the rocket, we need to break the problem into two phases:

Phase 1: Powered Ascent (with acceleration)

  • Initial velocity u=0u = 0 (since the rocket is fired from rest).
  • Acceleration a=20m/s2a = 20 \, \text{m/s}^2.
  • Time of powered flight t1=4.0secondst_1 = 4.0 \, \text{seconds}.

Using the kinematic equation for displacement under constant acceleration: s1=ut+12at12s_1 = ut + \frac{1}{2} a t_1^2 Substitute the values: s1=0×4+12×20×(4)2=160ms_1 = 0 \times 4 + \frac{1}{2} \times 20 \times (4)^2 = 160 \, \text{m} Thus, after the first phase, the rocket has risen 160 meters.

We can also calculate the velocity at the end of the powered ascent using: v1=u+at1=0+20×4=80m/sv_1 = u + at_1 = 0 + 20 \times 4 = 80 \, \text{m/s}

Phase 2: Coasting (no acceleration)

After the motor shuts off, the rocket continues to rise due to its upward velocity v1=80m/sv_1 = 80 \, \text{m/s}. In this phase, the only force acting on the rocket is gravity, which decelerates it at g=9.8m/s2g = 9.8 \, \text{m/s}^2.

The rocket reaches its maximum height when its velocity becomes zero. Using the equation: v22=v12+2as2v_2^2 = v_1^2 + 2a s_2 Here, v2=0m/sv_2 = 0 \, \text{m/s} (since the velocity is zero at maximum height), a=9.8m/s2a = -9.8 \, \text{m/s}^2, and s2s_2 is the additional height during the coasting phase.

Substitute the values: 0=(80)2+2×(9.8)×s20 = (80)^2 + 2 \times (-9.8) \times s_2 s2=64002×9.8=326.53ms_2 = \frac{6400}{2 \times 9.8} = 326.53 \, \text{m}

Total Maximum Elevation:

The total maximum elevation is the sum of the height from both phases: smax=s1+s2=160+326.53=486.53ms_{\text{max}} = s_1 + s_2 = 160 + 326.53 = 486.53 \, \text{m}

Thus, the rocket reaches a maximum elevation of approximately 487 meters.

Would you like more details or have any questions?

Here are five related questions to consider:

  1. How would the maximum elevation change if the rocket’s acceleration was higher?
  2. What happens if air resistance is no longer negligible?
  3. How long will the rocket take to fall back to the ground after reaching maximum height?
  4. How fast will the rocket be moving just before it hits the ground?
  5. What would the maximum height be if the rocket fired for a longer duration before coasting?

Tip: Always check units when performing calculations to avoid errors in your final answer.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Projectile Motion
Newtonian Mechanics

Formulas

s = ut + (1/2) a t²
v = u + at
v² = u² + 2as

Theorems

Kinematic Equations of Motion
Newton's Laws of Motion

Suitable Grade Level

Grades 10-12