(I) A stone is dropped from the top of a cliff. It is seen to ...

Question

(I) A stone is dropped from the top of a cliff. It is seen to hit the ground below after 3.25 s. How high is the cliff?

Accepted Answer

Hey, everyone in this problem, a child drops a ball from the top of a building which reaches the ground after 5.3 seconds. And we're asked to calculate the height of the building. We're given four answer choices all in meters. Option A 25.9 option B 2.75 multiplied by 10 to the exponent two. Option C 14 and option D 1.38 multiplied by 10 to the exponent two. So let's start by drawing a little diagram. So we have our child and they are going to drop this ball from the top of a building and that ball is going to travel all the way down until it reaches the ground. And we're gonna say that that distance that ball travels to the ground is H and H is the height of the building. OK. So H is what we are looking for now because all of the motion is downwards, we're gonna take downwards to be our positive direction. So we don't have to deal with any negative signs. You could do the opposite there or you just have to make sure that you account for that in all of your velocities and all of your accelerations. All right. So we have a kinematics type problem with this ball following from the top of this building. So let's go ahead and write out all the information we have. Now the initial velocity V knot is going to be zero. OK. We're told that this child drops the ball. So the ball is initially in this child's hand, it's not moving, they're gonna drop it. So that initial velocity is 0 m per second. The final velocity we don't, OK. Right. As the ball touches around, we don't know that speed, the acceleration is gonna be the acceleration due to gravity which is going to be positive 9.8 m per second squared. In this case. OK? Because we've chosen down to be a positive direction, the acceleration due to gravity works in the downward direction. So that is a positive value delta Y say that vertical displacement is going to be equal to the height. OK? That is what we are looking for. So that's the value we wanna calculate. And we're told that the time this takes is 5.3 seconds. All right. So in terms of our U AM or kinematic equations, if you call them that we have three known values V, not A and T, we wanna find delta Y in order to calculate the height. And so we can choose a U AM equation that has those four variables included. We have three known values. So we're good to go and we're gonna exclude the final velocity because that is not what we're looking for. And we don't have information about it. And that equation is going to be delta Y is equal to V not multiplied by T plus one half, multiplied by a multiplied by T squared. Substituting in our values, we get that the height H is equal to. While the first term on the right hand side, V not multiplied by T is just gonna be zero because V knot is zero. And so we're left with one half multiplied by the acceleration 9.8 m per second, slurred multiplied by the time 5.3 seconds, all squared. Now, in terms of units, we have meters per second squared multiplied by second squared. The unit of second squared is gonna divide out and we're gonna be left with just meters, which is exactly what we want for height. So that's good. Everything checks out there. And when we simplify everything on the right hand side, we get that the height H is gonna be approximately 137 0.6 or 1 m. OK? Now we wanna write this like the answer choices. So we're gonna write this with three significant digits. And what we have when we do that is that this is gonna be approximately equal to 1.38 multiplied by 10 to the exponent 2 m. And when we write it like this, we can see that this corresponds with answer choice D Thanks everyone for watching. I hope this video helped see you in the next one.

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(I) A stone is dropped from the top of a cliff. It is seen to hit the ground below after 3.25 s. How high is the cliff?

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