On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The free-fall acceleration on the moon is 1/6 of its value on earth. Suppose he hit the ball with a speed of 25 m/s at an angle 30 degrees above the horizontal. (b) For how much more time was the ball in flight?

Video transcript

Hey, everyone in this video, we're gonna work through the following problem. Consider an imaginary concept that the Olympics are held on another planet whose gravitational acceleration is 1/10 times its value on earth Deter. During the event of javelin throw, the spear is launched at an angle of 45° with respect to the horizontal and a takeoff speed of 20 m/s. We're asked to determine the difference between a spear's time of flight on the imaginary planet and it's time of flight on earth. We're given four answer choices, option a 31.8 seconds, Option B. 2.89 seconds, Option C 28.9 seconds And option D 26 seconds. OK. So for this problem, we have javelin throw. Let's draw a little diagram. We have our javelin and it's gonna be thrown at an angle of 45° from the horizontal At a speed of 20 m/s. OK. And we can break up that initial throw into the Y component and the X component. All right. So let's start writing the information we have for the other planet and for the earth. OK. We wanna know the difference between the time of flight on each of those planets. So let's just start by finding the time of flight on the other planet, the time of flight on earth and then we can find the difference. Now, for the other planet, yeah, we're told some information about its acceleration and the or the gravitational acceleration. OK. We know that the gravitational acceleration acts in the y direction or the vertical direction. So let's start with information in the Y direction and we know that the initial velocity and we're gonna use P for planet for the other planet. So we know that the initial velocity, The Y component is gonna be related to the sign. So we have 20 m/s Multiplied by sine of 45°. And we don't know the final velocity on the planet. We know that the acceleration is going to be 1/ gravitational acceleration. OK? So this is going to be 1/10 Multiplied by 9.8 m/s squared. And I've left this space in front because we need to choose directions. OK. We've said that the initial velocity is positive, it's going up into the right. So we're taking up into the right as our positive directions. That means that this gravity is going to act downwards in the negative direction. So the acceleration is going to be negative. Now, we know the displacement delta Y P is going to be zero m Ok. Why is it? zero m? Well, we're looking at the time of flight, right? Once this javelin stops flying, it's gonna have made its way back to the ground. And so the displacement is gonna be zero. OK? It's gonna be in the exact same y position as it was when it started and we don't know the time tea, but that is what we are interested in finding. Yeah, we're gonna call it T P for the time on the planet. OK. So we have three known values. One unknown. This is a straightforward kinematics problem. We're gonna choose the equation that includes V not P A P delta Y P, the three knowns and T P the unknown. And that equation is going to be the following. We have delta Y P is equal to V not P multiplied by T P plus one half AP multiplied by TP two. OK. Delta YP is zero m And this is equal to 20 m/s, multiplied by sine of 45° multiplied by T P was one half Multiplied by -1 10 multiplied by 9.8 m per second squared, Multiplied by TP two. OK. This looks really messy but we have one unknown here T P and the rest we just need to simplify. So if we simplify, we have zero on the left hand side, This is equal to 14.14, 2 meters per second multiplied by T P -0. meters per second squared Multiplied by TP two. Hey, we have this factor of T P in both of these terms. So we can simplify and factor that out. We have zero is equal to TP multiplied by 14.14, m/s minus 0.49 m per second squared multiplied by T P. OK. We had T P squared, we factored out A T P. So we're left with just T P. Yeah, we're trying to solve for T P. We have this factored equation equal to zero. This is just like when we're solving for the roots of an equation. So if either of these factors are equal to zero, then the left hand side will equal zero. OK? Something multiplied by zero will give us zero. So we can either have this first factor which is just T P equal to zero or we can have this second factor equal to zero. And that gives us 14.14, 2 meters per second minus 0.49 m per second squared multiplied by T P equal to zero. OK. If we solve for T P, we get that T P is equal to 14.142 m per second, divided by 0.49 m per second squared, which gives us a time of flight on the planet of 28.86 seconds. OK. So that's the time of flight on the other planet. Now, this zero second time we found as well. Hey, that makes sense because we are saying that we want to be at a displacement of zero m. We want the javelin to be on the ground. And before the flight, right at the very beginning, the javelin is going to be on the ground, ok? Before it gets thrown. And so it makes sense that at time zero, we also have this situation, OK? So 28.86 seconds is the time we were looking for for the planet. Now let's go back up and write out the information we know for earth. So for earth, remember we're trying to compare the times on the two planets. We're gonna look in the Y direction again and we're gonna use subscript E to indicate earth. So we have V not E and again, that same initial velocity 20 m/s multiplied by sine of 45°. We still don't know the final velocity. The acceleration is now just the regular acceleration due to gravity negative 9.8 m/s. Squared. Delta Y on earth is going to be the same zero m, ok? We want that javelin coming back to the earth and what we're looking for is T E that time. Ok. So we're gonna use, we have the exact same information pieces of information we're gonna use the exact same equation to calculate the time on the earth. So we have delta Y on the earth is equal to V not E multiplied by the time T E plus one half A E multiplied by T E squared. Subsequuting inner values, zero m is equal to 20 m/s. Multiplied by sine of 45° multiplied by T plus one half, multiplied by negative 9.8 m per second squared, multiplied by T E squared. OK. Now, this is the exact same situation that we had before. So we're gonna simplify and we're gonna factor and we're gonna do both of those things in one step this time since we know what we're doing now. OK. So we have zero is equal to, we can factor this term of T E O. In the first term, we're left with 20 m per second multiplied by sine of degrees. OK? That's gonna simplify to 14. m per second. From the second term, we have one half multiplied by negative 9.8 m per second squared. So we get minus 4.9 m per second squared in our um factor. And then we still have multiplied by T E can be factored at one T E but we still have another left in this second term. So same thing, we want either of these factors to be zero. If the first one is zero, the time is going to be zero seconds. OK? So that's at the very beginning before the javelin is thrown. The other time where this happens is gonna be 14.142 m per second minus 4.9 m per second. Squared. T E is equal to zero. We can solve for T E just like we solved for T P. It's gonna be 14.142 m per second, divided by 4.9 m per second squared. And we get a time on earth of 2.886 seconds. HM. And what you'll notice between these two times is on the other planet, we decreased the gravitational acceleration 10 times and it increased the time by 10. OK. So on Earth, we have 10 times more gravitational acceleration and the time is times less. All right. So this question, remember is asking for the difference of the time of flight. And so we're just gonna subtract the two, the time on the other planet minus the time on earth equal to 28.86 seconds minus 2.886 seconds. OK. Which gives us a difference of 25. seconds. And that is the difference that we were looking for if we go up to our answer choices and compare what we found And we're gonna round our answer choice and we see that this corresponds with answer choice. D the difference between the flight time on the imaginary planet and earth is approximately 26 seconds. Thanks everyone for watching. I hope this video helped see you in the next one.