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. (b) How much time will elapse after engine failure before the rocket comes crashing down to the launch pad, and how fast will it be moving just before it crashes?

Video transcript

Hello everyone. So in this problem, a spacecraft of mass 10,000 kg initially at rest is launched vertically with an outward acceleration of 10 m per second squared at an altitude of 500 m. The engine of the spacecraft cuts off what is the speed with which will be moving just before crashing down to Earth's surface. So we started by drawing a diagram for this problem. We have some spacecraft and it's initially arrest, so B. I. Is equal to zero m per second and it shoots up with an acceleration of 10 m per second squared to some Altitude of m. And the engine spacecraft cuts off, It's in free fall at this point, it will continue moving some maximum altitude and in this second this is the first phase of the problem. And then the second phase, It will reach this 500 m, but the same same velocity as it continues to move. So the important distinction to make is that the I in this second phase is equal to B. F one And from age equals 500 to back down, age equals zero. The spacecraft will have some final velocity, v. F two and we want to find this final velocity. You also recall that in free fall The acceleration is just the acceleration due to gravity, like at 9.8, m/s. Oh, and it's also important to note that this equation V I two is equal to V. F one. It's for the magnitudes if we want to consider direction, it would also be negative V. I two because since they're moving in different directions, So we can start off by writing what we would call as the dramatic equations. The first one is VF is equal to VI plus one is V F squared is equal to V I squared Plus two. A delta x one, delta X is equal to V I T plus one, a half a T squared. Now we want to find this, the F one as it will give us VI two. As all the equations with VF needs a V I. So let's since we have delta X, which is simply just delta age As the -500 m in the Phase two. So it's important to write that the delta H, It's equal to negative 500 m as it's falling down before phase one. Right. The second climatic equation, since we have this delta age as a V F one squared is equal to V I one squared plus two G delta H. And now we know that VI is simply zero And so V one is equal to the square roots of two times 9.8, m/s squared Times the m. We get the VF one Is 100 m/s. Now we can solve for this second phase final velocity. We can rewrite this expression since we're given delta age, this second cinematic equation as VF two squared is equal to V I two squared Plus two G Delta H. And now we can simply solve for BF two, which is the square root of V. I, too. You know, We know the I two as the F one, Which is this 100 m/s square plus two times 9.81 m 1st 2nd squared Times the negative. 500 m. F two, two, m/s. His answer choice, B 140.7 m/s. Hold this help have a great day.