A stone is catapulted at the top of a 75 m building as a...

A stone is catapulted at the top of a 75-m building. As a projectile, it is given an initial velocity of 20 m/s at an angle of 42". Answer the following in TWO decimal places. 1. At the launch point, the x-component of the velocity is blank m/s, while the y-component is blank m/s. 2. At the maximum height reached, the x-component of the velocity is blank m/s, while the y-component is blank m/s. 3. The time it takes to reach the ground is blank s. 4. At the landing point, the x-component of the velocity is blank m/s, while the y-component is blank m/s. 5. What is the "magnitude" of the velocity at the landing point then? 6. What is the "angle" that the velocity at the landing point makes with the horizontal? 7. What is the max height reached relative to the ground?

Transcript

00:01 In this video, i'm going to do a quick example looking at 2d kinematics, so a motion in two dimensions.

00:06 What we have is a projectile launched off the top of a building, and it has an initial velocity.

00:14 The initial equals 35 .2 meters per second, and it's launched at an angle theta equals 35 degrees above the horizontal.

00:27 And this projectile lands a horizontal distance d equals 125 meters away from the base of the building and we want to find the initial x and y components of velocity so v x v y initial i i want to find the height h of my building above the ground so how high was the projectile when it was launched i want to find the total time of my flight.

00:58 I want to find my final x and y components of velocity.

01:03 Vy f.

01:06 And i want to find my magnitude of my final velocity.

01:11 And i want to find the angle at which the projectile is going to strike the ground.

01:16 All right.

01:17 So let's start with our x and y components.

01:19 So here we are at some distance h above the ground.

01:24 Here's the ground.

01:25 We launch with an initial velocity v0.

01:28 At an angle of 35 degrees with our horizontal.

01:34 And our projectile after launch is going to travel up this way, reach its maximum point, and land somewhere over here at this distance d of 125 meters.

01:47 Okay, so to find the x and y components of velocity, we know that this angle is 35.

01:53 We know the magnitude of my velocity is 35 .2 meters per second.

01:58 So my initial x velocity is going to be 35 .2 times cosine of 35.

02:07 That's going to equal 28 .8 meters per second.

02:13 My initial y velocity is going to be 35 .2 times sine of 35.

02:22 And that gives me 20 .2 meters per second.

02:28 My initial velocity in component form is vo equals 28 .8x plus 20 .2 .2 meters per second.

02:44 Next, we want to find the time that this projectile is in flight, and i'm going to use the x direction motion to find this.

02:51 We know that in the x direction, we have no acceleration, so i'll have a constant velocity.

02:56 So i know my time is just given by time equals distance over velocity.

03:02 That's in the x direction.

03:03 That's my 125 meters divided by 28 .8 meters per second.

03:10 That gives me a time of 4 .34 seconds.

03:14 So from the time the projectile is launched to the time it hits the ground, 125 meters away from the base of the building.

03:21 That takes us 4 .34 seconds.

03:24 Next, i want to find the height of my building.

03:27 Okay, so i'm going to use this equation in the y direction.

03:30 So i know my y final position equals my y initial position, and this will be the height of our building, plus the initial times t, plus one -half a t squared, where here acceleration is going to be the acceleration due to gravity, and that's negative 9.

03:48 1 .81 meters per second squared.

03:51 I know my final position is zero...

Question

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