On level ground a shell is fired with an initial velocity of 40.0...

Question

On level ground a shell is fired with an initial velocity of 40.0 m/s at 60.0° above the horizontal and feels no appreciable air resistance. (b) How long does it take the shell to reach its highest point?

Accepted Answer

Hey everyone today, we're dealing with the problem about projectile motion, uniformly accelerated motion. So we're being told that a toy gun on a horizontal surface fires a bullet at 70°70° above the horizontal. If the bullet has an initial velocity of 39 m/s and Aaron resistance is negligible, we're being asked to find how long the bullet takes to reach the highest point. So how long does the bullet take? So before we do any of the math, let's just draw this out conceptually real quick. If we have something being launched at an angle, in this case a bullet from a toy gun, if this is the path of the motion and we have the horizontal here, the surface is being launched at then if it's being launched at an angle, then that means that not only is it moving horizontally in the X direction, it's also moving vertically in the Y direction. Excuse me, in the Y direction. And since we're being asked to find how long the bullet takes to reach the highest point, that means that we need to focus on the vertical motion. We need to focus on this Y aspect. So with that in mind, let's go ahead and write out our variables that we know and the ones that we need to find out. So we have the initial velocity which since we're dealing with vertical motion, we need to find the initial velocity of the Y component of the motion. We need to know the oral, we need to know the acceleration, we're trying to find what the time it takes to reach the highest point is. So that is our target. The final velocity is similar to the initial velocity in that we need to consider the initial velocity or the final velocity of the Y aspect only. We're not really concerned with the distance that it's being launched. We're only concerned with the height. So with that in mind. And the last thing would also be the displacement vertically, the height that it reaches and we're not giving that but we have three of the other values. Let's solve those out. So if we want to find the vertical component while we're given the initial velocity the initial velocity right here. But we need to multiply this by sign data. And the reason for this is because if we look back at our handy dandy diagram over here, our vertical component, If we look at this like a right triangle, like a right triangle. Are vertical component is opposite the angle, it is the opposite side of the angle, it is not adjacent. Like divert like the horizontal. So recalling our trig identities we have so kata. And since we're dealing with the opposite and the hypotenuse, we need to use sign. So that's why we use sign data over here. But solving that out, we have 39 m/s, multiplied by sine 70°, simplifying this and equating we get a value of 36.6 meters per second. Our acceleration. Again, we're concerned with vertical direction and to reach maximum height. We are acting against gravity, it is going up so it's acting against gravity. So our acceleration will be The negative force of gravity, so negative 9.8 m/s squared. We're trying to find our time and our initial and our final velocity at maximum height. Remember even though I've drawn it as a sort of right triangle to the left, a full uh projectile motion will follow a more parabolic structure. So at the highest point, let's draw this in green over here at the highest point is when the vertical velocity will actually read zero, it comes to a sort of standstill in vertical motion because it's gone up as much as it can. Now it's about to start its descent. So there is a period in time that maximum height when it sort of floats in the air. And we can see this if we throw a ball ourselves. But this means that the final vertical velocity at maximum height is zero m per second and we now have three of the five variables that we need. So we can go ahead and use a uniformly accelerated motion equation that excludes this vertical displacement. And that equation is that the final velocity is equal to the initial velocity plus the acceleration times the time. And again we're trying to find time. So simplifying this, let us sub in our values, we have zero and I'm leaving out the variables for the sake of simplicity and space zero is equal to 36.6 m/s Plus negative 9.8 m/s squared, multiplied by time in seconds. Simplifying this, we get negative 9.8 T. Is equal to -36.6. Simplifying this we get that T. Is equal to negative -36.6, Divided by -9.8. You gonna say final value of 3.74 seconds. Therefore the time it takes for the bullet to reach the highest point in his project. Ary with an initial velocity of 39 m per second and an angle of 70 degrees above the horizontal is option choice D 3. seconds. I hope this helps. And I look forward to seeing you all in the next one.

On level ground a shell is fired with an initial velocity of 40.0 m/s at 60.0° above the horizontal and feels no appreciable air resistance. (b) How long does it take the shell to reach its highest point?

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