As a science fair project, you want to launch an 800 g model rocket straight up and hit a horizontally moving target as it passes 30 m above the launch point. The rocket engine provides a constant thrust of 15.0 N. The target is approaching at a speed of 15 m/s. At what horizontal distance between the target and the rocket should you launch?

Question

Accepted Answer

Hi, everyone in this problem. We're being asked to calculate the horizontal distance separating the missile and the object. We'll have an unidentified object advancing toward a nuclear facility along the X direction at a constant speed of 40 m per seconds, located two kilometers above the ground level. We'll have a 15 kg missiles producing a thrust of Newton and oriented vertically upward, placed at ground level in the same X Y plane as the object, the missile is fired up and then hits the object. And we're being asked to calculate the horizontal distance separating the missile and the object at the instant of launching the missile. The options given are a 500 m b m C 1200 m and D 12 m. Awesome. So I am going to start off with listing out all the given information from our problem statement. So I am going to say that O is going to be the subscript for the object and M is going to be my subscript for the missile and for a collision to actually occur the missile and the moving object must be at the same position. So I'm gonna write down so that collision can occur then X O or the position of the object in the X direction and X M which is the position of the missile in the X direction have to be the same and Y O will also have to be the same as Y M or the position of the object. And the missile has to be the same in the Y direction as well. So the moving object has a constant speed and moves along the X direction, which I am going to say that that will be R V X O and that will be 40 m per second because we know that it is constant. Then we know that A X O is going to equals to zero m per second squared or the acceleration is going to be zero. The vertical position of the unidentified object is maintained at two kilometers above the ground level. So Y O Well equals to two km Or essentially YO will equals to 2000 m just like. So next uh the missiles initial position Y M I is going to be at the ground level which is zero m and then the miss the missiles in initial velocity, which is PM I in the Y direction is also going to be zero m per second which is starting from rest uh as I said, for collision to happen because we know that the Um vertical position of the unidentified object is going to be at two km above the ground level. Then the missile's final position at the time of collision is going to equals to that. So why MF will equals to 2000 m the missiles acceleration am Y in the right direction is still unknown. And when that is something that we're gonna be looking for and the time b when the missile reaches the moving object is also I know. So first what we wanted to uh start to be doing is to calculate the acceleration. So we will model the missile as a point like object. And we're going to apply Newton second law in the right direction. So applying Newton second law, we will get Sigma F Y which is in the Y direction well equals to the mass of the missile multiplied by the A Y or the acceleration also in the Y direction off the missile just like. So, So looking at the missile right here, the two different forces that will be acting upon our missiles are going to be first, the trust which is uh given to be 425 Newton from the problem statement. And the second is going to be the weight of the missile itself which is going to be m multiplied by G. So the Sigma F Y is then going to be F minus M multiplied by G equals to M multiplied by AM Y just like. So, so now we can actually rearrange this in order for us to get an equation for AM Y or the missiles acceleration in the Y direction. And that will then equals to B F minus M G divided by mass or A. We're going to substitute all of our known information. The force or the thrust is going to be 425 Newton. The mass is going to be 15 kg of the missile. The G is still 9.81 m per second squared. And all of that is going to be divided by kg. And that will give us the acceleration of the missile in the right direction to be 18.52 m per second squared, just like so awesome. So now that we found the acceleration, uh we can use that acceleration to calculate the time that is needed for the missile to actually reach the object using the kinematic equation. So using kinematic equation, we want to find the time that it takes for the missile from start to hitting the object. So the kinematic equation that we're gonna use is Y M F or the final position of the missile will equal to Y M I plus V M I initial Y multiplied by D plus half AM Y B squared. So looking at this in uh looking at this equation right here, we can actually simplify this directly. We know that the Y M I is going to be zero and V M I Y or the initial starting velocity is also going to be zero. So therefore, Y M F is only going to be half AM Y multiplied by T squared. Therefore, the equation for D can actually be obtained by the square root of two YMF divided by AM Y just like so awesome. So now that we get into this point, we can actually start substituting all of our information that we know. So D will then be the square root of Two multiplied by AYMF is going to be 2000 m. And am Y is the one that we just found, which is 18.52 m per second squared. And that will actually give us the time to be 14.7 seconds just like so awesome. So now that we find the time that it takes for the missile to go all the way to collision, we can actually use that time to find horizontal distance, separating the moving target and the missile or separating the unidentified object from the missile. That is going to also utilize the kinematic equation. So I'm gonna call this equ kinematic equation two and the previous one to be kinematic equation one. So the equation that we want to use is uh based on the x direction of the unidentified object. So X O F will equals to X O I plus V O T in the X direction plus half A O X T squared. However, we know that the uh identified object has a constant speed. So therefore, the acceleration will equals to zero. We can then simplify our equation. The one thing that we are interested at is actually X O F minus X O I, which is the difference in horizontal distance from the instant when the missile launches and instant when the missile actually collided with the identified object. So X O F minus X O I is the one that we are interested at and that will be equals to 30 X multiplied by T we have both V O X and T. So we can just plot those information in the V O X is going to be m per second, which is the constant speed that the unified object has in the X direction. And the time is 14.7 seconds, which is what we found previously. So that will actually give us a distance of 588 m, which is going to be the answer to this particular practice problem. So That will also correspond to option B in our answer choices. So option B is going to be the answer to this problem statement with a horizontal distance separating the missile and the object at the instant of the launch of the missile to be 588 m. So that will be all for this particular practice problem. If you guys still have any sort of confusion, please make sure to check out our adolescent videos on similar topic and that will be it. Thank you.

Verified Solution

Key Concepts

Newton's Second Law of Motion

Projectile Motion

Relative Motion