A football player punts the ball at a 45.0^∘ angle. Without an effect from the wind, the ball wo

A football player punts the ball at a 45.0^∘ angle. Without...

Key Concepts

Impact of External Forces on Projectile Motion

External forces, such as wind, can alter a projectile’s trajectory by affecting its velocity components mid-flight. When a gust of wind reduces the horizontal velocity, it directly influences the horizontal distance traveled, requiring adjustments in the calculations of the projectile’s overall displacement.

Range Calculation in Projectile Motion

The horizontal range of a projectile is determined by its initial speed, launch angle, and the acceleration due to gravity. The standard range formula assumes that the projectile lands at the same vertical level from which it was launched and is key for predicting how far the object will travel in the absence of additional forces.

Decomposition of Velocity into Horizontal and Vertical Components

In projectile motion, the initial velocity is split into horizontal and vertical components using trigonometric functions based on the launch angle. The horizontal component remains constant in the absence of horizontal forces, while the vertical component is affected by gravity. This decomposition is crucial for applying kinematic equations independently to each direction.

Projectile Motion

Projectile motion describes the motion of an object that is launched into the air and is subject to gravitational acceleration, with its horizontal and vertical motions treated independently. This concept is essential for understanding trajectories and calculating parameters such as time of flight, maximum height, and horizontal range.

Transcript

00:03 For this example, we're going to determine what the magnitude of the velocity of a football must have been, given that it traveled 60 meters horizontally and was punted at an angle of 45 degrees.

00:18 And then after that, what we're going to do is change up the problem a little bit.

00:24 And we're going to say that at the top of the football's trajectory, a gust of wind slows down the football by 1 .5 meters per second in the horizontal direction.

00:35 And then the ball is allowed to fall the rest of its trajectory.

00:40 And we're going to figure out how the range is affected after the ball gets slowed down up at the top here.

00:47 But just for the first part of the problem, we're going to act like there's no wind at all, and we just want to find what the initial speed must be.

00:57 So to do that, we're just going to start with our equation for the range when there's no air resistance, which is just b not squared times the sign, of 2 theta divided by g.

01:13 So we're just going to go ahead and solve this for v .0.

01:18 So i'll multiply both sides by g over sine theta.

01:21 I'm going to swap the sides around two.

01:23 So v .0 squared equals g times r divided by the sine of 2 times theta.

01:33 Now when we plug in theta equals 45 here, we should have that sign will be maximized because the argument to the sign function will be 2 times 45, which is 90.

01:50 So really we're just looking at the square root of gr in our situation here.

02:03 And if we calculate that, we get that the initial velocity is 24 .3 meters per second.

02:18 So the initial velocity must have been 24 .3 meters per second in order for the range to be 60 meters, given that we kicked it at a starting angle of 45 degrees.

02:29 Okay.

02:31 So now we're going to change up the problem a little bit.

02:33 And it's going to be modeled with what i've drawn here, where we're breaking up our projectile problem into two separate parts, where the first half is under normal conditions.

02:48 And the second half in blue is going to be under the conditions after the football gets pushed back by a gust of wind.

03:00 So when i talk about stuff from the first half, i'm going to use just our usual notation.

03:06 And when i talk about stuff from the second half, i'm going to use primes on everything, just to distinguish.

03:13 Okay.

03:15 So we want to figure out our new range.

03:17 And you can see that's going to be x2 from the diagram here.

03:23 And what we know is that x2 is just going to be equal to x1, right, our starting position.

03:34 Right here horizontally, plus whatever the new horizontal velocity is up top there after we get blown by a gust of wind.

03:46 So we'll call that v sub x prime times the amount of time it takes the ball to fall that second part of the trajectory, which we'll call t prime.

03:57 So we know x1 is just going to be equal to half of the range of our initial problem.

04:06 So x1 is just going to be our initial range over 2, which is 30 meters.

04:13 But we don't know v.

04:15 X prime or t prime yet.

04:18 Now lucky for us, we can figure those out without too much trouble.

04:22 First, let's look at v .6 prime...

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