Problem 19: A projectile is shot toward a hill that is d =...

Problem 19: A projectile is shot toward a hill that is d = 250 m away. The launch angle is ? = 45 degrees above horizontal with an initial speed of v0 = 70.1 m/s. The hill can be approximated as a plane sloped at ? = 18 degrees. Neglect air resistance. Part (a) Write an equation for y as a function of x, d, and ? for the line that defines the slope of the hill: y = tan(?) x - d tan(?) Part (b) Write an equation for y as a function of x, g (9.80 m/s^2), v0, and ? of the trajectory of the projectile: y = x tan(?) - (g x^2) / (2 v0^2 cos^2(?))

Transcript

00:03 In this problem, we have this inclined plane that represents some hill and we indicate this downward, this inclined downward direction x, and we measure the z -axis again from this inclined plane.

00:23 This angle from the vertical is given to be alpha and we have the origin right here.

00:29 Now we are throwing some object from this origin with some initial speed v0 that makes this angle theta with the z axis.

00:45 That will follow some pad and it will land on some point along this hill somewhere down the line.

00:56 We are asked to find this maximum maximum separation between the hill and this trajectory and we are also going to find this range along this here so let's call this first distance that max and the second one x max so we have this gravitation acceleration acceleration in the usual downward direction and since this this direction of this gravitational acceleration is not aligned with the you know the usual x -axis we are going to either rotate the frames or rotate this acceleration or for later we are going to separate into components so there are two ways apparently there are two ways to solve this problem and i will use the second one which is easier.

02:10 So i'm going to consider this gravitation acceleration and decompose it into components along the xan x -axis.

02:25 So like that and that.

02:29 So we have this angle of alpha here and the...

02:35 And along the x -axis we have this positive acceleration g times cosine alpha and along to z axis we have this negative okay there's not to put a negative sign here we have this negative acceleration g times sine of alpha it's negative because it is opposite to the positive x axis and now we have this constant acceleration motion in both x and z directions so i will simply write down my equations for that so i have x equal to x0 plus v0 x t plus one half a x t squared and for the z direction or for the z component of the position i have z zero plus v0 z t plus one half a z x times t squared.

03:52 So what are these values you know most of them? so we have x0 to be zero because it starts from origin and very similar.

04:03 We have z0 equal to 0 again we have the x component of the initial velocity to be v0 times okay so okay the thing is uh theta we have sine of theta and for the z component of the initial velocity we have v0 times cosine theta and for the acceleration for the x component we have plus g times cosine alpha and for the z component we have minus g times sine of alpha and we are also given that this theta is equal to 30 degrees alpha equal to 45 degrees v0 equal to 4 meters per second and the gravitation acceleration is 9 .8 meters per second squared.

05:17 Now in the first part we are going to find this maximum z value.

05:27 So the usual strategy is to take the time derivative of this z function which is explicitly given here and set it equal to the zero and solved for t or since the time derivative of the z gives us the z component of the velocity we have vz equal to zero z plus a z t so this z component of the velocity should vanish at this maximum point so this is what we have for for its condition okay, let us solve this for t...

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