We have no clue until we see the drawing. After all, the drawing shows its path.
Since the rocket’s acceleration is 3.00 m/s^3 * t, its acceleration is increasing at the rate of 3 m/s^3 each second. The equation for its velocity at a specific time is the integral of the acceleration equation.
<span>vf = vi + 1.5 * t^2, vi = 0 </span>
<span>vf = 1.5 * 10^2 = 150 m/s </span>
This is the rocket’s velocity at 10 seconds. The equation for its height at specific time is the integral velocity equation
<span>yf = yi + 0.5 * t^3, yi = 0 </span>
<span>yf = 0.5 * 10^3 = 500 meters </span>
<span>This is the rocket’s height at 10 seconds. </span>
<span>Part B </span>
<span>What is the speed of the rocket when it is 345 m above the surface of the earth? </span>
<span>Express your answer with the appropriate units. </span>
<span>Use the equation above to determine the time. </span>
<span>345 = 0.5 * t^3 </span>
<span>t^3 = 690 </span>
<span>t = 690^⅓ </span>
<span>This is approximately 8.837 seconds. Use the following equation to determine the velocity at this time. </span>
<span>v = 1.5 * t^2 = 1.5 * (690^⅓)^2 </span>
<span>This is approximately 117 m/s. </span>
<span>The graph of height versus time is the graph of a cubic function. The graph of velocity is a parabola. The graph of acceleration versus time is line. The slope of the line is the coefficient of t. This is a very different type of problem. For the acceleration to increase, the force must be increasing. To see what this feels like slowly push the accelerator pedal of a car to the floor. Just don’t do this so long that your car is speeding!!</span>
Answer:
the pe at the top of the building: 784 J
the pe halfway through the fall: 392 J
the pe just before hitting the ground: 784 J
Explanation:
Pls brainliest me
I had this question before
