(III) A toy rocket moving vertically upward passes by a 2.0-m-...

Question

(III) A toy rocket moving vertically upward passes by a 2.0-m-high window whose base is 8.0 m above the ground. The rocket takes 0.15 s to travel the 2.0 m height of the window. What was the launch speed of the rocket, and how high will it go? Assume the propellant is burned very quickly at blastoff.

Accepted Answer

Hello, everyone. Let's go through this practice problem. A tennis ball is hit vertically from the ground and passes by a 3 m tall flagpole. The ball takes 0.2 seconds to clear the entire height of the flagpole whose base starts 5 m above the ground. Calculate one, the speed of the ball after it got hit and two, the height reached by the ball option. A part 1 15.9 m per second and part 2 15 m. B 18.8 m per 2nd and 18 m C 15.9 m per 2nd and 18 m and d 18.8 m per 2nd and 15 m. OK. This is kind of a scary problem because there is quite a bit going on here. First off, we'll define the upward direction to be positive. And the problem gives us the most information about the period when the ball is traversing up at the height of the flagpole itself where we're told that the time for this period, I mean, you and we're gonna use the subscript one to label this period is 0.2 seconds and the displacement for this period or delta Y sub one is 3 m and we'll use U subscript one to refer to the speed of the ball when it reaches the base of the flagpole. Now, because we're analyzing the motion of an object under constant acceleration. This is going to be a prime example of where we want to use the kinematics equations of constant acceleration. Those three equations being that final speed is equal to initial speed plus acceleration multiplied by time displacement is equal to initial speed multiplied by time plus one half acceleration multiplied by T squared. And final speed squared is equal to initial speed squared plus two multiplied by acceleration multiplied by displacement. These are the three equations that we typically use in these types of problems. And the way we typically choose which equation to use is to choose the equation that includes the variable we're trying to solve for while also including variables that we know. So because part one of the problem is asking for the speed of the ball after it got hit, we are looking for the speed of the ball all the way from at the very bottom of the diagram, we don't have any information about the ball speed as it travels along the base. But the first thing we can do is find you sub one, let let's first try to solve for the speed of the ball by the time it reaches the bottom of the flagpole. So to find use of one we'll choose the equation that includes a speed while also including information about time and displacement which we were given for that period. And the equation that fits the bill is the second equation for displacement where we're go, we're going to use, use sub one for the initial time. So the displacement for that period is the 3 m and it's equal to the initial speed. We're trying to find use of one multiplied by the time. So 0.2 seconds multiplied by one half of the acceleration which is negative because it's in the downward direction. 9.81 m per second squared. That's just g multiplied by the square of the time which again is 0.2 seconds and don't forget to square it. So we're just going to algebraically solve this for U sub one by subtracting from both sides of the equation. The one half negative 9.81. So basically just make it positive. So U sub one multiplied by 0.2 seconds is equal to 3 m plus one half, 9.81 meters per second squared multiplied by 0.2 seconds squared. And then we divide both sides of the equation by 0.2 seconds. So we just solve for U sub one and find that it has a speed of 15.981 m per second. So this is the speed that the ball has when it reaches the base of the flagpole. But now we're gonna consider the motion of the ball from the point when it was hit to when it reached that base. So now we're concerned with this period, which I'm, I'm going to use the two subscript label to refer to that. So the top speed is a V sub two. However long it takes is T sub two. The displacement we're calling delta Y sub two, which is given in the diagram as being 5 m and the bottom speed, which is what part one is asking for is gonna be U sub two. Now, in this case, we have one speed and we're looking for the other speed at the other end of the motion. So this time, the third equation, the V squared equation is actually gonna work better for us because we have both speeds and we know the acceleration and we know the distance. So it's just the initial speed that we're trying to find. So our version of that equation is gonna be that V sub two squared is equal to U sub two squared plus two multiplied by the acceleration multiplied by delta yy two Y sub two. Remember that we're trying to solve for U sub two. So we're gonna subtract both sides of the equation by two A delta Y squared. So U sub two squared is equal to V sub two squared minus two A delta Y sub two. So now let's just plug in our values. So U sub two is equal to the square root of the sub two squared, which is a square of 15.981 m per second. Don't forget to square it minus two, multiplied by the acceleration. So negative 9.81 m per second squared multiplied by delta Y sub two which is 5 m, that's the length of the base. And if we put this into a calculator, then we find a speed about 18.8014 m per second, which can be rounded down to about 18.8 m per second. So that is our answer to the first part of the problem. And the second part of the problem asks us to find the height reached by the ball because for all we know the ball could have flown way higher than the flagpole's heights. So I'm gonna use the three subscript to refer to the ball's entire path where it's where it's final speed at the top of its motion is the sub three. The total distance is delta Y sub three, delta Y sub three being of course, what we're trying to find and its initial speed from the bottom, I'll call you sub three, which is of course, just going to be the same thing as U sub two, which we just found. So once again, we're gonna use the same kinematics equation that we used at the end of part one of the problem where the final speed is V sub three squared, it's equal to the initial speed. U sub three squared plus two multiplied by the acceleration multiplied by the displacement. Which is what we're trying to find now because we're now looking at the point where the ball reaches its maximum speed. It's important to understand that its maximum height, it's gonna have a speed of zero because that's the point where the ball is finally reached rest where it's slowed down. So zero is equal to U sub three squared. As I just mentioned, that's just gonna be the same thing as our answer to the previous part of the problem. So you sub two squared plus two multiplied by the gravitational acceleration. G multiplied by delta Y sub three, which is what we want. So to solve this equation for delta Y sub three, we're gonna subtract U sub two squared from both sides of the equation and divide both sides by two G. So we find that delta Y sub three is equal to negative U sub two squared divided by two G. Now let's put this into a calculator that's negative the square of U sub two. So 18.8014 m per second squared divided by two multiplied by the gravitational acceleration of negative 8.1 m per second squared. If we put this into a calculator, then we find a distance of about 18 m. So that is our answer to the second part of the problem. So our answers to the problem are 18.8 m per 2nd and 18 m which corresponds to option B. So that is our answer and that is it for this problem. I hope this video helped you out and please check out our other tutoring videos which will give you more experience with these types of problems. That's all for now and I hope you have a lovely day. Bye bye.

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