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(1I) A projectile is fired with an initial speed of 46.6 $\mathrm{m} / \mathrm{s}$ atan angle of $42.2^{\circ}$ above the horizontal on a long flat firing range. Determine $(a)$ the maximum height reached by the projectile, (b) the total time in the air, $(c)$ the total horizontal distance covered (that is, the range), and (d) the velocity of the projectile 1.50 s after firing.

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a) Maximum height $=50$ metersb) Total time is 6.39 secondsc) Horizontal Distance $=221$ metersd) $v_{\text {total}}=38.3 \mathrm{m} / \mathrm{s}$ and $\theta=25.7^{\circ}$

Physics 101 Mechanics

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

Motion Along a Straight Line

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supper party. Let's find the maximum height. We know that the Y final square, uh, equals V. Why initial squared plus two times G Delta y on this case are why initial is going to be zero. So we can say that here, why Max would be equal to the why final squared minus the why initial squared, divided by two times G and this is equaling negative. The initial squared sine square of clear initial divided by negative to G. And so this would equal be an initial squared sine squared of data initial divided by two G for the negative signs cancel out and we can solve. So this would be 46.6 meters per second. Ah, quantity squared times sine squared of 42 0.2 degrees, all divided by two times 9.80 meters per second squared. This is giving us 50 0.0 meters for part B. They want the total time in the air so we can stay. Why final equals. Why initial quest B y initial tp was 1/2 gt squared again. Here downwards is positive. So we can say that tea is simply gonna equal to two times being initial sign of Fae, The initial divided by G but solved two times 46.6 meters per second, multiplied by sine um of 40 2.2 degrees. My apologies. Divided by G So 9.80 meters per second squared. He is found to be 6.39 seconds and of course, zero seconds. However, this dancer does not apply, of course. And so, um, we can say that see is found to be Oh, we want to find the total horizontal distance covered So Delta X displacement in the extraction with equal v x times t and this would equal 46.6 meters per second coastline of the angle fully 2.2 degrees and then divided by 9.80 meters per second squared. And this is giving us 6.39 seconds. So this would be my my apologies. This would be this would 46.6 coast on a 42 and not not divided by G, but rather times 6.39 seconds. The time that we just found here. My apologies. Delta X would then equal, of course, 221 meters and finally for teeth. We want to find the, um essentially the final velocity. Um, we know that the, uh, the velocity of the projectile 1.5 seconds after firing is going to just simply be the vector. Some of the horizontal velocity and the vertical velocity s o the, um, for the horizontal VX initial, which equals V X final. Because there isn't any acceleration in the extraction, this would simply be 46 0.6 meters per second co sign of 42 0.2 degrees. This is giving us 34.5 meters per second. And then we can say that the wife final equals B. Why? Initial question, G t, um, we can say that this is gonna equal 46.6 sign of 42.2 degrees, minus 9.80 meters per second squared multiplied by here, it will be 1.5 seconds, and this is equaling 16.6 meters per second. So let's draw. It's right in vector notation. We can say that the final velocity, at 1.5 seconds would be equal two 34.5 I, um, plus 16.6 j hat and then the units, of course, would be meters per second. Let's find the magnitude of the final velocity. After 1.5 seconds, this would be 34.5 squared plus 16 0.6 squared all to the 1/2 power. This is giving us 38.3 meters per second and to find the angle we simply take are tan of the wide component divided by the X component. And this is giving us 25.7 degrees. We have a positive why in a positive X. So this would be in the first quadrant, essentially so 25.7 degrees above positive X axis. So this would be the direction and the magnitude of the velocity again after 1.5 seconds that it's the end of the solution. Thank you for watching

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