A Rocket initially at rest, launched vertically from a ground with an acceleration of 20 m/s2 for 6.0s until its engine stops. Disregarding any air resistance, calculate what is the maximum height above the ground the rocket achieved? ( Hint: after the stop of engine the rocket continues to move under acceleration of gravity)
a.
0.79km
b.
1.9km
c.
0.8km
d.
0.73km
e.
1.1km
The correct answer and explanation is:
Let’s break down the problem step by step:
Step 1: Analyze the rocket’s motion during the engine-powered phase (0 to 6 seconds)
The rocket starts from rest and accelerates vertically with a constant acceleration of 20 m/s² for 6.0 seconds. We can calculate the velocity at the end of this phase using the equation: v=u+atv = u + at
Where:
- uu is the initial velocity, which is 0 m/s (since the rocket starts from rest),
- aa is the acceleration, 20 m/s²,
- tt is the time, 6.0 seconds.
v=0+(20)(6)=120 m/sv = 0 + (20)(6) = 120 \, \text{m/s}
So, the velocity of the rocket when the engine stops is 120 m/s.
Step 2: Calculate the height at the end of the engine-powered phase
We can find the displacement (height) during the acceleration phase using the following equation: s=ut+12at2s = ut + \frac{1}{2} a t^2
Substitute the values: s=0×6+12×20×62=0+0.5×20×36=360 ms = 0 \times 6 + \frac{1}{2} \times 20 \times 6^2 = 0 + 0.5 \times 20 \times 36 = 360 \, \text{m}
So, the rocket travels 360 meters during the 6 seconds of engine-powered flight.
Step 3: Analyze the motion after the engine stops (6 seconds onward)
Once the engine stops, the rocket moves upward under the influence of gravity alone, so the acceleration is now −9.8 m/s2-9.8 \, \text{m/s}^2. The velocity at the start of this phase is 120 m/s, and the rocket will continue to rise until its velocity reaches zero.
Using the equation v2=u2+2asv^2 = u^2 + 2as, where:
- v=0v = 0 m/s (final velocity when it stops),
- u=120u = 120 m/s (initial velocity at engine stop),
- a=−9.8a = -9.8 m/s² (acceleration due to gravity),
- ss is the displacement during this phase.
0=1202+2(−9.8)s0 = 120^2 + 2(-9.8)s 0=14400−19.6s0 = 14400 – 19.6s 19.6s=1440019.6s = 14400 s=1440019.6=734.69 ms = \frac{14400}{19.6} = 734.69 \, \text{m}
So, the rocket will rise 734.69 meters after the engine stops.
Step 4: Total maximum height
The total maximum height is the sum of the height during the powered phase and the height during the coasting phase: Total height=360 m+734.69 m=1094.69 m≈1.1 km\text{Total height} = 360 \, \text{m} + 734.69 \, \text{m} = 1094.69 \, \text{m} \approx 1.1 \, \text{km}
Final Answer:
The maximum height the rocket achieves is approximately 1.1 km, which corresponds to option e.