A 7500-kg rocket blasts off vertically from the launch pad with a...

Question

A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25 m/s2 and feels no appreciable air resistance. When it has reached a height of 525 m, its engines suddenly fail; the only force acting on it is now gravity. (a) What is the maximum height this rocket will reach above the launch pad?

Accepted Answer

Hello everyone. A rocket of mass 10,000 kg initially at rest is fired vertically with an upper acceleration of 10 m per second squared at an altitude of 500 m, the engine of the rocket cuts off what is the maximum altitude it achieves? We can begin by drawing a diagram to better visualize this problem. So we have a rocket and it's launched with an upward acceleration of 10 m per second squared And we know that is initially arrest BI zero m for a second. It reaches some height. H equals 500 m. So we know that delta H is equal to 500 minus zero, which is just 500 m now it continues with some velocity keeps moving. This is not the maximum hype. So we have some VF and we can split it into these two phases is one And we will also have a Phase two we'll reach some maximum height that we want to find age is equal to H max the delta H in this second phase two, H max minus 500. Now it is also important to know at this H max The F two which is different from the F one is equal to zero meters per second and is no longer moving. That is the maximum height. Now an important distinction to make into this problem is that the final velocity in this first phase, Equal to the initial velocity in the 2nd phase. All right too, so now we have all the steps begin solving this problem, recall the three came back equations As VF is equal to BI was 80. The second one V F squared is equal to B I squared Plus two A Delta X. And the 3rd 1 which is Delta X. is equal to the IT. Plus one half a t squared. So now when we look at what values were given, were given in this first regime delta H B I and A. And we want to find VF one. So we can get V I two to solve for V F two. As all of the equations involving VF must also have a V I. So we know we can either use one or two. Now we don't have time. So he wants to use this second cinematic equation. So for the first engine right there, second kinetic equation is V squared is equal to V I one squared plus two G dot H. You know, that is initially arrest. So the I one is equal to zero. To be readily solved for V F one, which is just the square root of two. G delta H. Oh, actually it is not G is an upward acceleration of a which is 10, it is only G in a second regime as it is in free fall on the second regime. So using 10. Now we get that is to the square root of two times m per second squared age, 500 m in first regime And we get that VF one is equal to 100 m/s, which we also know two. Now we can set up a this same kinetic equation, this second regime BF two squared is equal to B I two squared Plus two. Now it is G as it's in freefall delta H. And we want to solve for something in delta H. We want silver H. Max so we can rearrange this equation for delta age. You get that delta H Go to VF two squared minus E I two squared over two G. And Delta H. As we recall is H. -500. We can continue to solve for H max as VF two squared minus V I two squared over two G plus 500. And we can make the substitution of the values V. F two Squared As we know at this maximum height, the F two is 0. So this cancels and we're just left with negative V. I two squared which we solved. And his first section the negative V. I. Or you can make the substitution now 100 m/s squared over two times negative 9.81 m per second squared Plus the 500. Finally we get that H max. It was approximately equal to 10:10 needles answer choice. Hey 10 10 m. Have a great day

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