A. $$34.6 mathrmm / mathrms$$B. $$=3.53 mathrms$$C. $$61.2 mathrmm$$D. $$=141 mathrmm$$E. $$-9.80 mathrmm / mathrms^2$$




You are watching: How long does it take the shell to reach its highest point?

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19:02


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Chris D.



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Video Transcript

for this trouble, on the topic of motion in 2 dimensions, were told that a shell is fired from level ground at 60 levels above the horizontal, at an initial velocity of 40 m per second. If it feels no air resistance, we want to find the horizontal and also vertical components of its initial velocity the time it takes to reach its greatest suggest, its maximum height above the ground. It's horizontal distance once it lands and its horizontal and also vertical components of acceleration velocity at its greatest suggest. So firstly, we want to discover its horizontal component of initial velocity V, not X, and also we're not. X is equal to we not and also the speak to authorize of the firing angle alpha naught. And this is 40 meters per second Times the coal sine of the angle above the horizontal, which is 60°.. And so the horizontal component of initial velocity Is 20 m/s. The vertical component of the initial velocity V, not Y, is equal to the not sign of alpha naught. And this is its initial velocity 40 meters per second. And the white component is multiplied by this authorize of 60 levels. And so this gives us the y component to be 34 0.6 meters the second. So we have actually the initial velocity damaged up right into its X and Y components. Next for component B. We desire to discover the time it takes to reach its highest possible point. Oh, this object will certainly relocate in a parabola and at its greatest suggest Its vertical component of velocity will certainly be zero. And so we have the right to use the sort of magic equation V. Y is equal to the note why plus A. Y. Times T. And we can rearrange and also resolve for the time T. So T. Is equal to the last rate, V. Y minus the initial speed in the Y direction Z. Not why over the acceleration in the Y direction A. Y. Now the acceleration is that only as a result of gravity. So this is zero minus 34 .6 m/s separated by the acceleration. Which is that due to gravity minus 9.8 m per square second. Since we take downward to be negative, this offers us a time of 3.53 secs. So that's the moment it takes for the shower to reach its greatest allude. For part C. We desire to find the maximum elevation of the shall around the ground. Now aget make it is a kind of magic equation. This time Vienna twice squared or B. Y squared rather is equal to V. Not Y squared plus to a Y delta Y. Which is why minus? Why not? And so to solve for the Y displacement? Why manners? Why not? This is equal to V. Y squared minus V. Not y squared separated by to a Y. That's a speed at the greatest suggest of zero- the initial speed in the Y direction. 34.6 meters per second squared Over two right into -9.8 m/km2. So calculating we acquire the maximum elevation over the ground off the shell To be 61.2 meters next for component D. We want to calculate the horizontal distance that we shall go, go as soon as it lands. So the complete time in the air is twice the moment to the maximum height. And so we can usage the kind of magic equation X minus X. Keep in mind is equal to v. Not X plus a fifty percent a X T squared, but tright here is no acceleration, the X direction considering that the velocity is constant. So this is equal to the initial horizontal component of velocity 20 meters per second. So actually we recognize we're not X times T plus a half a XT square. We've foracquired the t Times twice the time to the maximum elevation, which is 2 times 3.53 seconds. And so we acquire the maximum horizontal distance covered by the shell to be 100 And 41 m. And ultimately for part E, we want to discover the horizontal and vertical components of its acceleration and velocity at the greatest point. Now, at maximum elevation we have V. X. To sindicate BV narcotics, considering that the horizontal component of velocity remains continuous throughout the journey And this is 20 m/s. At the maximum height V. Y is equal to zero considering that it momentarily concerns rest at all points in the activity, there's no acceleration in the horizontal direction and the Vertical component of acceleration is consistent and also that is -9.8 meters per square second. Or that because of gravity.