Show that, for a given initial speed, the horizontal range of a projectile is the same for launc

Show that, for a given initial speed, the horizontal range...

Key Concepts

Projectile Motion

Projectile motion refers to the two-dimensional motion of an object that is launched into the air and moves under the influence of gravity. This concept involves breaking the motion into independent horizontal and vertical components, with the horizontal component having constant velocity and the vertical component undergoing constant acceleration due to gravity.

Horizontal Range

The horizontal range is the total horizontal distance traveled by the projectile from its launch point until it returns to the same vertical level. It is determined by the initial speed, the angle of projection, and the acceleration due to gravity.

Range Formula

The range of a projectile can be calculated using the formula R = (v² sin(2?)) / g, where v is the initial speed, ? is the angle of projection, and g is the acceleration due to gravity. This formula encapsulates how the horizontal distance depends on the launch angle through the sine of double the angle.

Trigonometric Identities and Symmetry

A key insight in solving problems involving different launch angles is the use of trigonometric identities. For example, the identity sin(90° + x) = cos(x) shows that for launch angles 45° + ? and 45° - ?, the value of sin(2?) becomes the same because sin(90° ± 2?) ultimately equates to cos(2?). This symmetry in the sine function ensures that, for a given initial speed, the horizontal range remains identical for these two complementary angles.

Transcript

00:01 For this exercise, we have to consider two project tiles that were thrown with the same initial speed, but at different angles.

00:09 So the first one was thrown at an angle of 45 degrees plus alpha, and the second one was thrown at an angle of 45 degrees minus alpha.

00:21 What the exercise asks us to do is to show that the horizontal range of the first one, that i'm going to call d is the same as the horizontal range of the second one.

00:37 Okay, so both of them are g.

00:40 So let's calculate it.

00:43 Well, in order to calculate the horizontal range of each of them, let's first consider a projectile thrown at a certain angle theta, and let's calculate how much time it takes for it to hit the ground.

01:00 So we're going to use the equation for y, the y coordinate, that's equal to y0 plus v0y.

01:11 That's since the angle is theta, v0y is v0 times sine theta times t minus gt squared over two.

01:26 The minus here is because the gravity acceleration is pointing downward.

01:34 When the project tile hits the ground again, y0 equals y, we can simply disregard these two terms.

01:45 They cancel out.

01:46 Both of them are zero.

01:49 And then we can solve this equation here for t.

01:56 So t equals 2 v0 sine theta over g.

02:04 So that's the time when the project tile hits the ground again.

02:10 Okay, so to carry on, we have to calculate d.

02:16 D is the horizontal range.

02:19 That's simply going to be the horizontal velocity v0, cosine theta, times the time it takes for the project tile to hit the ground, the time it travels in the air.

02:33 And that is the time we found before...

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