A projectile is launched from the ground at an angle of 45^∘. What initial speed must the projec

A projectile is launched from the ground at an angle of...

Key Concepts

Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration. In the context of projectile motion, these equations are used separately in the horizontal direction (where acceleration is zero) and the vertical direction (where acceleration is due to gravity). They allow us to relate time, displacement, velocity, and acceleration, providing a framework for solving for unknown variables in motion problems.

Simultaneous Equations

Many projectile problems involve finding more than one unknown, such as the initial speed and the time of flight. By setting up equations for both the horizontal displacement and the vertical displacement and solving them together as a system, simultaneous equations enable us to determine these unknowns. This method is fundamental in connecting the independent motions to arrive at a comprehensive solution.

Projectile Motion

Projectile motion refers to the two-dimensional motion of an object launched into the air, which moves under the influence of gravity while neglecting air resistance. Its motion can be separated into horizontal and vertical components, which are analyzed independently. This concept is crucial for understanding how different forces affect motion in different directions.

Decomposition of Initial Velocity

This concept involves breaking the initial velocity of an object into horizontal and vertical components using trigonometric functions, typically cosine for the horizontal component and sine for the vertical component. Decomposing the velocity is essential because it allows for the independent application of kinematic equations to each component, which is necessary for solving projectile motion problems.

Transcript

00:01 For this problem, we are told that a projectile is launched from the ground at an angle of 45 degrees.

00:05 We are then asked, what initial speed must the projectile have in order to hit the top of a 120 meter tower located 180 meters away? so to begin, we know that our acceleration vector, a of t, well, we're not undergoing any acceleration in the x -axis.

00:23 So it would be zero in the x -axis.

00:25 And if we're going in meters, it would be negative 9 .8 meters per second.

00:30 Squared in the y axis, which then means that our velocity v of t would have to be equal to some constant in the x -axis, then negative 9 .8t plus a constant in the y -axis.

00:46 We have that the projectile is launched from the ground at an angle of 45 degrees, so we'd also have that essentially the relative angle that's going on here, we could write this as being equal to the magnitude of v times cos, or this would have to be equal to, i should say, magnitude of v times, oh, actually, i'm getting ahead of myself here.

01:13 What we'd have, excuse me, what we'd have to have is that based on the fact that we were being launched at an angle of 45 degrees, that would mean that our initial velocity in the x component would have to be equal to the initial velocity in the y component.

01:32 So we'd have to have that v of zero equals, well, c1, c2, clearly.

01:42 And we have that c1 must be equal to c2, but we don't have anything else that we can immediately say just yet.

01:49 So all we can say is that c1 equals c2, and i'll call that c2 just for now.

01:54 So c1 equals c2 equals c.

01:57 So we have v of t must be equal to c minus 9 .8 t.

02:05 Then we'd have that r of t must be equal to c t plus i'll call it now c1 and we'd then have ct minus 9 .8 t squared over 2 plus a new c2 and we have again, now r of 0 would have to be equal to 0 .0.

02:32 We're starting from 0.

02:34 So that means that when we evaluate this at, or when we evaluate the r of t that we found when t equals 0, we'd have this would be c1, c2.

02:45 So we have c1 equals c2, and they both equal 0.

02:50 So r of t then will just be equal to ct minus 9 .8t squared over 2.

02:59 Or, excuse me, let me correct myself, it's ct, ct minus 9 .8t squared over 2...

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