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8. An object is projected from (he ground, in (o the air; with an initial velocity of (80i + 160f)m/s How far will be its position relative to the point of projecti...

Question

8. An object is projected from (he ground, in (o the air; with an initial velocity of (80i + 160f)m/s How far will be its position relative to the point of projection afler 2Oseconds?

8. An object is projected from (he ground, in (o the air; with an initial velocity of (80i + 160f)m/s How far will be its position relative to the point of projection afler 2Oseconds?



Answers

A projectile is fired vertically from Earth's surface with an initial speed of 10 $\mathrm{km} / \mathrm{s}$ . Neglecting air drag, how far above the surface of Farth will it go?

For an object which is thrown vertically upward with the initial velocity V. Note and the acceleration acting on the objectives towards the downward direction and its magnitude G. So the position of the object with respect to its initial position is the initial velocity V not into time, T -1 of cheese into the square. Now let's substitute the values. So why is we notice 28 meters pose again into the time. T in the time, tease. One second minus half of Gs 9.81 m per second squared into one second square. The position of the object after 1 2nd days. Yeah, 23.1 m. So after one seconds the object will move 23.1 m upward.

For an object which is launched vertically upward with an initial velocity V. Nord. The acceleration of the of the objectives towards the downward direction and that the G So we can break the final velocity squared at the object is the initial velocity squared -2. Key into the change in the vertical displacement, delta Y. And then this delta Y will become max that it's at max maximum height. Uh the final velocity will become zero. So we can eliminate this and from this we can write H max is cuba into the data way and that is a cure into were not square upon to G. Now let's substitute the values the maximum height of duties, The square of the initial velocity that is 28 meters per second squared upon two and two G. That is 9.81 m per second squared at the maximum height of duties 39.959 m are in three significant figures. The maximum height is 40 me this wait, right.

So for this question will need to use the fact that the kinetic energy is equal to half times the mass, times the velocity of the object squared, and its potential energy is given by the negative of the gravitational constant, multiplied by the two masses divided by the distance between them. So what we can do first is set up an equation due to the fact that energy must be conserved in the situation. So on the left hand side of the equation here, we have the initial energy of the state, which we know must equal the final energy of the state. The initial energy is made up of the kinetic energy here, given by half Mv squared summed up with the initial potential energy in the final state there is no kinetic energy and all of the energy is stored as potential energy because this is a point when the projectile has reached its highest point and a stationary, so there is no kinetic energy is, the velocity is equal to zero. So all of the energy is the potential energy. Know that the D. Here is actually equal to the height above the surface of the earth. Which is what we're trying to obtain minus the radius of the earth. So we are given these values here and we can sub these into this expression in order to obtain a value for D. What we can do first to make this easier is cancel these ems out. So what we're left with is a half times the initial velocity squared which is 10,000 m per second. Yeah squared minus the gravitational constant which is 6.67 times 10 to the minus 11. Multiplied by the mass. Which is given here 5.98 times 10 to the 24 over the radius of 6.37 times 10 to the six is equal to again the gravitational constant times this mass again. Mhm. Yeah. Mhm Over D sobbing in these numbers and rearranging for D. We can then obtain that D. Is equal 23.2 times 10 to the seven m. As D is our only unknown in this entire expression. We then know that D. Is equal to hey H minus R. And R has a value of 6.37 times 10 to the six m. So in order to work out page, we have to do D plus are which is this value we've obtained plus the radius. So this gives us a final answer for the height above the surface of 2.5 times 10 to the seven m. Okay. Which is what we were trying to get in the question.

To solve this problem. I am doing the diagram first. Just look at it carefully. After that. I will simplify the problem. So the diagram looks something like this which I am growing here. I will also So all the points that are acting on it. So let this point is the this point HP. This is why this is point B. Death. This angle is 30 degrees. This angle it 60 degree. This is you. This point is okay. So here the arrangement in client plane is equal to O. P. So I can write the value of range is equal to who you is squared by G multiplication. Gosh! Cheetah! Same. He to minus alpha. Bye courses square alpha On simplification, I can devalue R is equal to do you squad by G multiplication cost 60 degree signed. 30 degree by plus is glad 30 degree which is equal to do you square by three G. So of Cindy is correct here. So also be is correct in this problem.


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