A projectile is launched from point O with the initial...

A projectile is launched from point $O$ with the initial conditions shown. Determine the impact coordinates for the projectile if $(a) v_{0}=60 \mathrm{ft} / \mathrm{sec}$ and $\theta=$ $40^{\circ}$ and $(b) v_{0}=85 \mathrm{ft} / \mathrm{sec}$ and $\theta=15^{\circ}$

Key Concepts

Range Calculation

The range of a projectile is the total horizontal distance traveled from launch to impact. It is obtained by multiplying the horizontal component of the constant velocity by the time of flight, linking the concepts of vector decomposition and time of flight in projectile motion.

Time of Flight

Time of flight is the total duration that the projectile remains in the air. It can be determined by analyzing the vertical motion, specifically by calculating the time taken for the projectile to rise and fall under gravity until it reaches the impact level.

Projectile Motion

Projectile motion describes the trajectory of an object that is launched into the air and is affected only by gravity (assuming air resistance is negligible). This concept involves analyzing the object’s motion in two dimensions, where the horizontal motion is constant and the vertical motion is uniformly accelerated due to gravity.

Vector Decomposition

Vector decomposition involves breaking down the initial velocity into horizontal and vertical components using trigonometric functions. This is crucial for projectile motion analysis because it allows the independent study of the motion in the horizontal and vertical directions.

Transcript

00:01 We've got two scenarios, each with the same setup.

00:03 So for a, we're given our vo and our theta, vo and our theta.

00:09 And let's go ahead and start solving for a.

00:14 So the equations of motion will be x equals vot times the cosine of theta and y equal t times the sign of theta minus gt squared y x.

00:34 Okay, to find the impact point, we're going to check the position of the projectile at some horizontal positions.

00:58 So we've got 60 feet and 40 degrees.

01:03 So let's do y at 80.

01:05 Let's figure out where why is it easy.

01:07 At y at 80 will be equal to 80 times the tangent of 40 minus g, which is 32 .174 times 80.

01:23 Squared over 2 times 60 squared times the cosine squared of 40 and that will equal 18 .4.

01:38 The ball's still flying.

01:47 So right here it'll be at like 16 .4.

01:52 Then our next point let's check at why equals 160 or 140, excuse me, 140 which is right here and we can figure our y for 140 will be equal 140 times the tangent of 40 times or minus 32 .174 times 8, 140 squared.

02:30 174 times, let's get rid of that whole thing there, 140 squared divided by 2 times 60 squared times the cosine squared of 40.

02:46 This is equal negative 31 .774.

02:52 Okay.

02:54 So our first impact point is going to be at 140 and negative 31 .774.

03:08 And both of these will be in feet.

03:10 So it's going to be, i'm going to make a different color here.

03:13 I'm going to make it this color.

03:15 So it'll be somewhere in here.

03:21 Okay.

03:22 Then we're going to do the same thing, except i'm going to switch to, i think, a pink.

03:28 And remember, for b, we have 85 feet per second and 15 degrees.

03:36 So let's do again, y at 80 will equal 80 times this time the tangent of 15 minus 32 .174 times 80 divided by 2 times 85.