The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0° above the horizontal, some of the tiny critters have reached a maximum height of 58.7 cm above the level ground. (See Nature, Vol. 424, July 31, 2003, p. 509.) (a) What was the takeoff speed for such a leap?

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Hey everyone. So today we're dealing with the problem about projectile motion and uniformly accelerated motion. So the question tells us that we have an animal, grasshoppers that can jump As high as 95.2 cm. So let's highlight that. That's very important, 95.2 cm above the ground. So that is their maximum height. And we're being asked to find the initial velocity Of a grasshopper, the takeoff speed, given that it leaps at 54° above the horizontal. So let's think about this conceptually really quickly. We're if a object or an animal in this case jumps at an angle to the ground, then that means that no longer only has a horizontal aspect of motion, it has the vertical aspect as well. That's just labeled these as X and Y. And we're dealing with the maximum height given the maximum height. And we're being asked to determine the takeoff speed, the initial velocity using height. So we need to use the vertical aspect of our motion of grasshoppers motion to go ahead and solve this. So let's go ahead and write out what we know well, the initial velocity in the Y direction, because we're dealing with height is what we're solving for this is our target. Let me write that a little nicer. This is our target. What else do we have? Well, we're not given time, Time has been completely excluded but were given a few other key factors were given that the change height, the maximum height that it can attain is 95.2 cm. And let's just convert this to meters because we deal with the sybase units. So to convert that for call that for every one m we have 10 to the second centimeters centimeters. So that means we have 0.952 m 0.952 m. Our acceleration because the grasshoppers jumping, it's acting against gravity, it is going up. So the acceleration is therefore the negative force of gravity or negative 9.8 m/s squared. And we're only using maximum height to figure out our initial velocity. And we know that at maximum height oops not an initial it's final at maximum height is when an object sort of comes to a standstill because remember even though I've drawn it as a triangle here, projectile motion has a more parabolic shape. So at the highest point is when the vertical velocity comes sort of to a stance to it because it's at that point that the direction changes, it sort of halts and air for a moment before coming back down. And it's a phenomenon that we can see if we throw a ball ourselves. So that means that the initial or final velocity, the final vertical velocity is zero m per second. So we have our three values. We have three of the five variables we need. So let's use an equation that has these three variables but doesn't include time because we're not given time and it's very difficult to try and solve for it and it would be unnecessary in our case, so the equation we can use for this is that the final velocity squared is equal to the initial velocity squared plus two A two times the force of acceleration multiplied by the displacement. The change in vertical displacement here. So substituting in the values that we do know we have zero squared and I'm leaving out the units for the sake of simplicity and space is equal to v naught squared and let me just write on our wise here because those are important Plus two times negative 9. Multiplied by Delta Y, which is 0.952 m. Simplifying this, We get that zero zero is equal to uh the initial velocity squared minus 18.66. Therefore The initial velocity squared is equal to 18.66. So we square root both sides. And we get that The initial velocity is equal to 4.32 m/s, 4.3, m/s. Therefore we have the initial velocity in the vertical direction. But we're still not done. We need to find the initial velocity as a whole. We're not just looking for the horizontal component. So what we need to do, well, we can look at our trigger metric functions to do this. We have our vertical velocity of course, but we need to remember that the vertical velocity is actually part of the tree geometric functions If we write out our tree geometric identities that we can recall Socotra, the vertical component lies opposite the angle and if we consider this to be a right triangle, that means it lies opposite to the angle and it is adjacent to the hypotenuse in a sense. But since we're opposite to the angle we can go ahead and use sign. And we know that the initial velocity initial velocity just as general, including both vertical and horizontal components, is equal to the initial or the vertical velocity. My bad. Oh sorry, what am I saying? The vertical component is simply the initial velocity that includes both the horizontal and vertical components. Multiplied by sine data again because of our triggered a metric identity here. So we know what the vertical component is now. It is 4. m/s. 4.32 m/s. And if that is equal to the initial velocity times sine data than the initial velocity, what we're looking for initial velocity is equal to 4.3 to 4. m/s, divided by divided by Sign of 54°, Which will give us a final answer of 5. meters per second. And this is our true true final answer. So the initial velocity it takes for a grasshopper to leap At a height of 95.2 cm above the ground at an angle of 54°. His answer choice C 5.34 m/s. I hope this helps, and I look forward to seeing you all in the next one.

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Key Concepts

Projectile Motion

Kinematic Equations

Trigonometric Functions