A rocket is fired straight up through the atmosphere from the South Pole, burning out at an altitude of 208 km when traveling at 5.8 km/s.
(a) What maximum distance from Earth's surface does it travel before falling back to Earth?
I know how to do the work, but i can't get the answer. I need the answer.
To find the maximum distance from Earth's surface that the rocket travels before falling back, we can use the concept of time of flight.
We know the rocket burns out at an altitude of 208 km. Let's assume this point as the highest point reached by the rocket.
First, we need to calculate the time taken to reach the highest point:
Time taken = Distance / Speed
Here, the distance is the altitude reached, which is 208 km (or 208,000 meters), and the speed is given as 5.8 km/s (or 5,800 meters/second).
Time taken = 208,000 meters / 5,800 meters/second
Calculating this value will give us the time taken to reach the highest point.
Next, since the rocket is fired straight up and its motion is symmetrical, it will take the same amount of time to fall back to Earth. Therefore, the total time of flight is twice the time taken to reach the highest point.
Total time of flight = 2 * Time taken
Now, we can calculate the maximum distance from Earth's surface:
Maximum distance = Speed * Total time of flight
Using the value of the speed given as 5.8 km/s, and the calculated value of the total time of flight, we can calculate the maximum distance.
Maximum distance = 5.8 km/s * Total time of flight
Now, substitute the calculated value of the total time of flight into the equation and calculate the maximum distance to get the answer.