A rocket lifts of the surface of the Earth with a constant acceleration of How f

Question

A rocket lifts of the surface of the Earth with a constant acceleration of How fast will the rocket be going 1 minute later? How far will it have travelled in that time? In 1971, Apollo 15 astronaut David Scott dropped a hammer and a feather on the moon to demonstrate that in a vacuum all bodies fall with the same constant acceleration. If he dropped them from rest from about 4 feet above the ground, how long did it take them to fall? Use the fact that the constant acceleration due to the moon's gravity is roughly - 5.2 Evaluate each of the following indefinite integrals by using the given substitution. (a) integral 2(2x + 4)^5, u = 2x + 4 (b) integral 4x^3/(x^4 + 1)^2, u = x^4 + 1 (c) integral (1 + squareroot x)^1/3/squareroot x, u = 1 + squareroot x (d) integral sec 2t tan 2t dt, u = 2t (e) integral 9r^2 dr/squareroot 1 - r^2, u = 1 - r^2 (f) integral squareroot x sin^2 (x^3/2 - 1) dx, u = x^3/2 - 1

Explanation / Answer

Solution:15

Initial Velocity of rocket is u = 0 since it is moving from rest

It has constant acceleration of 20m/sec2

acceleration due to gravitity = 9.8 m/sec2

therefore net acceleation = 20-9.8 = 10.2 m/sec2

Velocity after 1 minute = 60 sec

Apply newton first law of motion

v = u +at

v = 0 + 10.2 * 60= 612 m/sec

Distance = S = ut + (1/2) at^2

                     = 0 + (1/2) 10.2 * 60^2

                    = 18360 m

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