(5%) Problem 9: A robotic cheetah can jump over obstacles....

(5%) Problem 9: A robotic cheetah can jump over obstacles. Suppose the launch speed is v0 = 4.81 m/s, and the launch angle is ? = 46.2° above the horizontal Part (a) What is the maximum height, h, in terms of v0, ?, and g? Part (b) What is the maximum height, in meters? Part (c) Given the same launch speed, what launch angle, in degrees, would yield a maximum height that is 40.5% greater than the height in Part

Transcript

00:01 In this video, i'm going to be looking at projectile motion.

00:04 So what we have is a robotic cheetah that's designed to jump over obstacles.

00:09 Okay, and it jumps with an initial velocity of the initial equals 4 .81 meters per second.

00:17 And this cheetah jumps at an angle of theta equals 46 .2 degrees relative to the horizontal.

00:26 And what we want to find is a formula to find the maximum.

00:30 Height that this cheetah reaches, okay, in terms of initial launch velocity.

00:37 So the initial and our launch angle and our acceleration, which is going to be the acceleration due to gravity, and that equals negative 9 .81 meters per second squared.

00:50 Once we found that formula, we want to put in our numbers and find the actual numerical value of the maximum height that the cheetah reaches.

01:01 So i'll call that h1.

01:03 Okay.

01:04 And then once we found that value, we want to find the value of the launch angle.

01:10 If the cheetah were to reach a maximum height of 40 .5 % greater than our first height that we found.

01:19 Okay.

01:19 So i'll call that h2 equals 1 .405 times h1.

01:28 Okay.

01:29 So what i'm going to use to find that initial h1 is the equation v final squared equals v initial squared plus two times acceleration times height.

01:42 That will be my total vertical distance.

01:45 I know when i reach my maximum height, my velocity will be zero.

01:49 So this becomes zero equals the initial squared plus two times acceleration times height.

01:58 And we're looking at purely the vertical direction here.

02:02 So we need to find our initial vertical velocity in terms of our initial velocity.

02:09 So that's v sub -zero here and our launch angle.

02:14 And that equals the initial velocity times the sign of the launch angle.

02:20 So now i have zero equals v initial squared times.

02:28 Sine squared of my launch angle.

02:32 And now we're going to have minus 2g.

02:35 See, i've replaced acceleration by our value for gravity.

02:40 That's negative 9 .81 meters per second squared times height.

02:45 And i can rearrange this equation to get height, and i'll put the subscript 1 in there.

02:52 It's going to be equal to v initial squared, sine squared of that launch angle divided by 2g...

Question

Please give Ace some feedback