I- The Arc tan of four The Arctan of Four A projectile is...

I- The Arc tan of four The Arctan of Four A projectile is fired from the ground at an initial speed of $v_o$. At what angle should it be fired so that its horizontal \"range\" is precisely the same (number of meters or yards or whatever) as its peak height? \"Range\" refers to the horizontal displacement traversed by the projectile by the time has returned to the ground. HINTS: 1. Derive an equation that expresses the dependent variable $x$ ("range") as a function of the independent variable $\theta$. This equation will also make reference to the constants $g$ and $v_o$. 2. Independently, derive an equation that expresses the dependent variable $y$ ("pea height") as a function of the independent variable $\theta$. This equation will also make reference to the constants $g$ and $v_o$. It might start off by having time as an independent variable, but you've got to get rid of that somehow. It's not given i the problem.

Transcript

00:01 In this video, i'm going to be deriving the range equation.

00:05 And what the range equation is is an equation that gives us the total horizontal distance that a projectile travels.

00:13 Okay.

00:13 This is given in terms of our initial velocity and our launch angle.

00:18 Okay.

00:19 So we have a projectile.

00:21 It's launched at some angle.

00:25 Okay, theta.

00:27 Okay, so i'll call this the horizontal.

00:30 This is theta.

00:31 Range is going to be from where the projectile is launched to where it lands back on the ground, and that range is r.

00:39 So we're going to be asked several leading questions to come up with this equation for range, and the answer that we'll get will be r equals v initial squared.

00:52 Okay, so that's my initial launch velocity, v.

00:55 Initial, times the sign of two times our launch angle, k divided by g, where g is the acceleration due to gravity.

01:06 So my first question is we want to find an equation that gives the horizontal position of this object or projectile as a function of time.

01:14 I know there's no, for this problem we're neglecting error resistance.

01:18 So there's no forces acting in the horizontal direction.

01:22 The velocity is going to be constant.

01:24 Okay, so my horizontal position, and i'm going to call this r, and that's going to equal, the x component of my initial velocity times the total time the projectiles in the air.

01:36 Okay, that x component of velocity is just my total, the magnitude of my velocity vector times cosine of the launch angle.

01:47 Okay, so times t, and that gives me my total horizontal distance.

01:51 Now we're asked if the total time that the object stays in the air is dependent on what's happening in the x or in the y direction.

01:59 I already know that we have no forces acting in the horizontal direction.

02:03 So that's not going to give us very much information.

02:06 We do have a force and that's due to earth's gravitational field acting in the vertical direction.

02:11 Okay.

02:12 So that's going to affect the time it takes us to come back down to the ground after being launched.

02:18 Okay...

Question

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