A rock is thrown from the top of a 35 m building directly...

A rock is thrown from the top of a 35 m building directly downwards with an initial speed of 5 m/s. At the same time, a cannon on the ground fires a projectile directly upwards at 30 m/s.a. At what height do the rock and the projectile pass each other? b. Does the projectile make it to the top of the building? c. How much time does the rock take before falling to the ground? USE KINEMATICS

Transcript

00:01 Okay, so we're throwing a rock down with a speed of 5 meters per second down, so that's negative 5 meters per second, from a building that is a height of 35 meters.

00:09 And we're also firing a cannonball up with a speed of 30 meters per second.

00:13 We want to know at what point will they pass each other.

00:17 So we can use our position kinematic equation, so y equals y -0 plus v -t, t, plus 1⁄2, at t squared, and apply this to both the rock and the cannonball, or the projectile.

00:31 So for the rock, we're going to have y1 is equal to h, the height of the building, which, let's just plug in the numbers there.

00:42 So 35, minus five times time, and then plus one -half gt squared, or rather minus one -half g -t -squared, because we're going to be accelerating downward.

00:56 So let's change that to minus gt squared.

01:00 And there we go.

01:03 And now we can write a similar equation for the projectile, except its initial height is zero from the ground, and its initial speed is positive 30, and then that minus one -half gt squared as well.

01:18 And the point at which they are at the same height is when these two values are the same.

01:23 Y1 equals y2.

01:24 So let's set these two equations equal to each other.

01:27 So we have 35 minus 5t, minus one half gt squared on one side.

01:36 And then on the other side, we have 30t minus one half gt squared.

01:43 Now you can tell there is a one half gt squared on both sides.

01:48 So that will cancel out.

01:50 And our equations will then become 35 minus 5t is equal to 30t.

01:59 And combining like terms, adding the 5t to both sides, we get 35t on 1 .2.

02:04 Side and 35 on the other for the meters, which means they will pass at a time of one second.

02:13 So it'll take one second for them to reach the same height.

02:16 Now we can plug that time into either of our initial equations to find what that height will be.

02:24 So let's go ahead and plug it into this first one.

02:26 So we have 35 minus 5 times 1, minus 1 1⁄2 times 9 .8 times 1 squared.

02:33 And we get a height, of 25 .1 meters.

02:39 And you can plug that into the second equation and just verify that that is also the case.

02:44 And that proves that we have found the correct height.

02:47 So this is the height at which they will pass each other.

02:52 Now part b asks if the projectile will make it to the top of the building.

02:56 And we can tell that just by finding the maximum height.

03:00 And so we can use our time independent kinematic equation...

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