A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 75.6 m/s² for 1.96 seconds, at which point its fuel abruptly runs out. Air resistance has no effect on its flight. What maximum altitude (above the ground) will the rocket reach?
The Correct Answer and Explanation is :
To determine the maximum altitude reached by the model rocket, we need to analyze its motion during two distinct phases: the powered ascent (while the engine provides thrust) and the unpowered ascent (after the fuel is exhausted and the rocket coasts upward under the influence of gravity).
Phase 1: Powered Ascent
During this phase, the rocket starts from rest and accelerates upward with a constant acceleration.
Given data:
- Acceleration, ( a = 75.6 \, \text{m/s}^2 )
- Time under power, ( t = 1.96 \, \text{s} )
- Final velocity at the end of powered ascent (( v_f )): Using the equation of motion: [ v_f = v_i + a \cdot t ] where ( v_i ) is the initial velocity (0 m/s since the rocket starts from rest), we get: [ v_f = 0 + 75.6 \times 1.96 = 148.176 \, \text{m/s} ]
- Altitude gained during powered ascent (( h_1 )): Using the equation: [ h_1 = v_i \cdot t + \frac{1}{2} a \cdot t^2 ] Substituting the known values: [ h_1 = 0 + \frac{1}{2} \times 75.6 \times (1.96)^2 = 145.21 \, \text{m} ]
Phase 2: Unpowered Ascent
After the fuel is exhausted, the rocket continues to ascend due to its inertia until gravity decelerates it to a stop.
- Altitude gained during unpowered ascent (( h_2 )): Using the kinematic equation: [ v_f^2 = v_i^2 + 2 \cdot a \cdot d ] Here, ( v_f = 0 ) (velocity at the peak), ( v_i = 148.176 \, \text{m/s} ), and ( a = -9.8 \, \text{m/s}^2 ) (acceleration due to gravity, negative because it’s acting downward). Solving for ( d ) (which is ( h_2 )): [ 0 = (148.176)^2 + 2 \times (-9.8) \times h_2 ] [ h_2 = \frac{(148.176)^2}{2 \times 9.8} = 1120.21 \, \text{m} ]
Total Maximum Altitude (( H )):
The total altitude is the sum of the altitudes from both phases:
[ H = h_1 + h_2 = 145.21 \, \text{m} + 1120.21 \, \text{m} = 1265.42 \, \text{m} ]
Therefore, the rocket reaches a maximum altitude of approximately 1265.42 meters above the ground.
This problem illustrates the application of kinematic equations to analyze motion under constant acceleration, first under engine thrust and then under gravity alone.