18 - Free Fall Motion Problems in Physics (Acceleration due to Gravity), Part 7

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Күн бұрын

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In this lesson, we consider a problem that involves a rocket during powered flight moving with velocity upwards, in opposition to the acceleration due to gravity.
To solve this type of problem, we use the same equations of motion, but we must break up the problem in to two parts. The first part is calculating the position and velocity of the rocket before engine cuttoff. After that, the problem simplifies to a rocket traveling upwards to an apex in motion influenced only by gravitational acceleration.

Пікірлер: 47

@ebenezertawiah1486 3 жыл бұрын

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@MrBelkhou Жыл бұрын

Hi everyone, question can also be answered using free fall equation with zero initial velocity. The height is 308.2 m = (1/2) * g * t² and find t = 7.9 s to get down to the ground. Then add the other time t = 8.52 s to reach maximum height which gives the total time of flight 16.42 s.

@Vr1_touseef 2 жыл бұрын

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@masaamri1172 5 жыл бұрын

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@kevinv546 4 жыл бұрын

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@MathAndScience 4 жыл бұрын

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@MathAndScience Жыл бұрын

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@gabrielmchilomo4723 3 жыл бұрын

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@Meerfs 3 жыл бұрын

For c) can't we just use Y=Y0+V0t+1/2at^2 ? We know the variables and get Time from here and not calculate Velocity, that we are not asked for?

@carultch Жыл бұрын

Acceleration is not constant in this problem. There's a period when the rocket accelerates from its thrusters, and a period when the rocket coasts in free fall. You have to put these two periods together as a piecewise problem of two constant acceleration periods, to solve for the maximum height.

@drball2558 5 жыл бұрын

Nu complicated rocket designs to ascend a new electron into the ionosphere

@GideonBoateng-ez2oo 3 ай бұрын

Why was the height negative

@Miles2Achieve 4 жыл бұрын

acceleration is 2m/s2 and gravity acceleration is 9.8 which is pulling the rocket downwards , why we are not doing 9.8-2 for resultant acceleration or 2 is the final acceleration considering 9.8 ?

@ZikyFranky 3 жыл бұрын

Gravity acceleration is (-9.81)

@carultch Жыл бұрын

It is given in this problem that the acceleration is 2 m/s^2 upward. How this is caused, is irrelevant to the problem. The thrusters will provide the upward force needed to both oppose gravity AND cause an acceleration of 2 m/s^2. So on net, the force per unit payload mass of the rocket, is 11.8 N/kg.

@TriviaQuizocity 3 жыл бұрын

How do you aply this to gravity with respect to starting distance to find the velocity at the ending destance. This equasion should not include time, just distance from object. I can not use the equasion “square root of 2gh” because that is implying that the stringth of gravity is constant and does not change with respect to distance so that equasion does not work for me.

@carultch Жыл бұрын

Near the Earth's surface, gravity might as well be uniform, and the equation ultimately turns into v=sqrt(2*g*h) when falling a distance of h. When h is in meters, and the radius of Earth is in megameters, the scale of h relative to the radius of Earth is insignificant, and the gravitational field is close enough to uniform that it makes no difference to account for gravity's variation with position. If you wanted to account for how gravity varies with position, then you'd use the equation GPE = -G*M*m/r, and construct two versions of this with r=R and r=R+h in each of them. So you'd get deltaGPE = -G*M*m*(1/(R+h) - 1/R, which reduces to deltaGPE=G*M*m*(1/R - 1/(R+h)). Equate to KE=1/2*m*v^2, the little m cancels, and we get v=sqrt(G*M*(1/R - 1/(R+h))). G= universal constant of gravitation M = mass of Earth R = initial radial position from center of Earth, not necessarily the radius of Earth.

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I run my own site www.MathAndScience.com. Thanks! Jason

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@drball2558 5 жыл бұрын

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