(II) Piles of snow on slippery roofs can become dangerous proj...

Question

(II) Piles of snow on slippery roofs can become dangerous projectiles as they melt. Consider a chunk of snow at the ridge of a roof with a slope of 24°.

(c) If the edge of the roof is 10.0 m above ground, how far out does the snow hit the ground?

Accepted Answer

Hey, everyone. Welcome back in this problem. We're told that rocks on a slippery mountain could turn into swift projectiles as they slide for us to imagine a rock at the ridge of a mountain incline 22 degree angle below the horizontal with an initial velocity of 6 m per second. The mountain's edge is 13 m above the ground. And we're asked to calculate the horizontal distance between the base of the mountain, the point at which the rock hits the ground. Ok. So we're looking for the horizontal distance between the base of the mountain from the point where the rock hits the ground. We have four answer choices all in meters. Option A 4.2 option B 5.5 option C 7.9 and option D 9.6. So let's go ahead and draw out what we have going on and we have this cliff and we have this slope uh coming downwards that makes an angle of 22 degrees with the horizontal. And our projectile or rock is gonna be on that edge and it is moving with an initial velocity of 6 m per second. It is gonna come off of that cliff, off that mountain edge and it is gonna land somewhere on the ground. And we're looking for that distance, that horizontal distance that it travels between the base of the cliff and it's landing spot that is going to be our distance d that we are looking for. Now, let's go ahead and write out all the variables we have acting here. So in the x direction, in the horizontal direction, what do we have? Well, we have our initial velocity and our initial velocity, let's take downwards into the right to be positive. OK. So the directions of motion, so our initial velocity is going to be positive and it's gonna be 6 m per second multiplied by cosine of that angle 22 degrees. OK. So we're taking the horizontal component of that initial velocity of 6 m per second delta X that displacement or the distance traveled in the X direction. That's what we're looking for. That's that devalue that we're looking for. And we don't know how long it takes before it hits the ground. So this is everything we have going on in the extraction and we're assuming that we have no air resistance. And so we have no acceleration in that horizontal direction. So our initial velocity is gonna be the same as our final velocity in that extraction. OK. So if we can figure out the time that it takes, then we'll be able to calculate this distance D that we want right now, in order to calculate that time, we can think about what's happening in the vertical or y direction. Because the time it takes this rock to hit the ground, whether we're looking in the horizontal direction or the vertical direction is the same, it's the same amount of time. It's the same rock landing in one place. So if we think about the Y direction, what do we have going on? Well, the initial velocity we can find just like we did in the horizontal direction, we have 6 m per second multiplied by sun of 22 degrees. In this case, because we're taking that vertical component. Now the final velocity when this rock hits the ground, we don't know what that is. We know that this is going to have a displacement, a vertical displacement of 13 m because the mountain's edge is 13 m above the ground. So that rock is gonna be traveling 13 m downward, downwards are positive direction. So that displacement is positive, we have acceleration due to gravity. It's gonna be 9.8 m per second squared. Again, that's positive because downwards is our positive direction in this case. And we don't know how long this takes, but we do know that it's the same amount of time as in the horizontal direction. So what we can see now is that we have three known values in the Y direction. We have V not delta Y and A, we can use those with a kinematic or U AM equation to calculate T. So let's go ahead and do that first. So we are looking for this T value, we're gonna choose the kinematic equation that has those four variables we just mentioned in it. And that equation is going to be delta Y is equal to V. Not why multiplied by T one plus one half multiplied by A Y multiplied by ty squared. Substituting in our values 13 m is equal to 6 m per second, multiplied by sign of 22 degrees multiplied by T plus one half, multiplied by 9.8 m per second squared multiplied by T squared. And we have a T squared term A T term in a constant term. We have a quadratic equation here. We're gonna move everything to one side so that we have it set equal to zero. That will allow us to use a quadratic formula to solve for T. When we do that, we get zero is equal to 4.9 m per second squared multiplied by T square. They were simplifying these coefficients plus 6 m per second, multiplied by sign of 22 degrees multiplied by T minus 13 m. And we need to identify our A, our B and our C values. So in this case, our A value is gonna be 4.9 m per second. Squared B is gonna be 6 m per second multiplied by S of 22 degrees and C is gonna be negative 13 m. As I mentioned, we're gonna be using the quadratic formula to solve the quadratic formula tells us that in this case T is going to be equal to negative B plus or minus the square root of B squared minus four AC all divided by two A substituting in the A B and C that we identified T is going to be equal to negative 6 m per second. Multiplied by sign of 22 degrees plus or minus the square root of 6 m per second multiplied by sign of 22 degrees, all squared minus four, multiplied by 4.9 m per second squared, multiplied by negative 13 m. And all of this is divided by two multiplied by 4.9 m per second squared. All right, this looks very messy but we can go ahead and use a calculator and we're gonna get two values for T one when we use plus one, when we use minus in that numerator. And the two values we get T is equal to 1.41 554 seconds approximately. Or we get T is negative 1.87 424, a second. Now we're talking about time here. So having a negative value does not make sense. So we can eliminate that second option. And we find that we have this time T of about 1.41554 seconds. So that's how long this rock is in, in the air before it hits the ground. If we go back up and remind ourselves what we're looking for, we're looking for this horizontal distance between the base of the mountain and the point where the rock hits the ground. We have only one known value V in that horizontal direction. So we need to, to find t first, we now know that this time T is equal to 1.41 554 seconds. So we can use that information to calculate that missing distance D. Now, now remember that in the horizontal direction, we're assuming we have no air resistance, we have no acceleration. And so all we have is that the speed V or the velocity is delta X displacement divided by the time delta T, OK. In this case, our speed 6 m per second multiplied by cosine of 22 degrees. That is equal to the distance D you were looking for divided by the time which we know is 1.41 554 seconds. We just calculated that. So if we multiply 1.41554 seconds on both sides, we get that, that distance D is going to be equal to about 7.87 or 8 m. And that is the distance from the base of our cliff that the rock is going to hit the ground. If we go up to the answer choices we were given and we round to two significant digits. We can see that the correct answer corresponds with option C 7.9 m. Thanks everyone for watching. I hope this video see you in the next one.

Struggling with Physics?

(II) Piles of snow on slippery roofs can become dangerous projectiles as they melt. Consider a chunk of snow at the ridge of a roof with a slope of 24°.
(c) If the edge of the roof is 10.0 m above ground, how far out does the snow hit the ground?

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Struggling with Physics?

(II) Piles of snow on slippery roofs can become dangerous projectiles as they melt. Consider a chunk of snow at the ridge of a roof with a slope of 24°.(c) If the edge of the roof is 10.0 m above ground, how far out does the snow hit the ground?

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(II) Piles of snow on slippery roofs can become dangerous projectiles as they melt. Consider a chunk of snow at the ridge of a roof with a slope of 24°.
(c) If the edge of the roof is 10.0 m above ground, how far out does the snow hit the ground?