The range R of a projectile is defined as the magnitude of...

The range $R$ of a projectile is defined as the magnitude of the horizontal displacement of the projectile when it returns to its original altitude. (In other words, the range is the distance between the launch point and the impact point on flat ground.) A projectile is launched at $t=0$ with initial speed $v_{i}$ at an angle $\theta$ above the horizontal. (a) Find the time $t$ at which the projectile returns to its original altitude. (b) Show that the range is $R=\frac{v_{\mathrm{i}}^{2} \sin 2 \theta}{g}$ [Hint: Use the trigonometric identity $\sin 2 \theta=2 \sin \theta \cos \theta .]$ (c) What value of $\theta$ gives the maximum range? What is this maximum range?

The range $R$ of a projectile is defined as the magnitude of the horizontal displacement of the projectile when it returns to its original altitude. (In other words, the range is the distance between the launch point and the impact point on flat ground.) A projectile is launched at $t=0$ with initial speed $v_{i}$ at an angle $\theta$ above the horizontal. (a) Find the time $t$ at which the projectile returns to its original altitude. (b) Show that the range is $R=\frac{v_{\mathrm{i}}^{2} \sin 2 \theta}{g}$
[Hint: Use the trigonometric identity $\sin 2 \theta=2 \sin \theta \cos \theta .]$
(c) What value of $\theta$ gives the maximum range? What is this maximum range?

Key Concepts

Projectile Motion

Projectile motion refers to the motion of an object that is launched into the air and is subject only to the acceleration due to gravity, assuming air resistance is negligible. It involves a two-dimensional movement that can be analyzed by separating the horizontal and vertical components, which makes it easier to apply the principles of kinematics separately in each direction.

Time of Flight

The time of flight in projectile motion is the total time that a projectile remains in the air. It is determined by analyzing the vertical component of the motion, where the projectile, after being launched upward, returns to its original altitude. By setting the vertical displacement to zero and solving the corresponding kinematic equation, one can find the time it spends in the air.

Range Formula

The range of a projectile is the horizontal distance it travels during its flight. It is obtained by multiplying the horizontal component of the velocity by the time of flight. The derivation of the range formula involves using the time of flight determined from the vertical motion and then applying the horizontal kinematic equation to yield an expression involving the initial speed, launch angle, and gravitational acceleration.

Trigonometric Identities

Trigonometric identities, such as the double-angle identity sin 2? = 2 sin ? cos ?, are used to simplify expressions in projectile motion problems. These identities allow for the conversion of expressions involving products of trigonometric functions into a single trigonometric function, which simplifies the derivation and interpretation of formulas like the range of a projectile.

Optimization in Projectile Motion

Optimization in projectile motion involves finding the conditions that maximize a particular outcome, such as the range. This typically involves taking the derivative of the range expression with respect to the launch angle and setting it to zero to find the critical points. It is well-known that the maximum range is achieved when the launch angle is 45°, which maximizes the contribution of both the horizontal and vertical components.

Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration, and are fundamental to analyzing projectile motion. These equations allow us to determine key parameters such as displacement, velocity, and time based on the initial conditions and constant gravitational acceleration, which is crucial for solving problems involving the motion of projectiles.

Transcript

00:01 Okay, so this is a simple third radical question where we have to find first the amount of time it takes this ball this projectile to go from its initial spot to its end spot where they're the same altitude we have the constant acceleration of gravity and it's negative 9 .81 meters per second squared or we'll replace 9 .81 g so it's just negative g really and then we have that so in order to find out the amount of time let's split this sort of path into two sections we have going up and we have going down.

00:31 Okay, let's do going down to blue.

00:32 Now the thing to know is that going up, the green path, and going down, the blue path actually takes the exact same line of time.

00:39 So all we need to do is calculate the green path and multiply it by two.

00:42 So how long is the green path going to take going up to the peak? well, we know that at the peak, the vertical velocity is going to be zero.

00:49 That's just what happens at the peak, right? so let's use this formula.

00:52 Vy equals to v in initial y plus vertical acceleration times time so let's deconstruct us we know that v y is going to be zero we know that v i y using you know this triangle is going to be equal to v i sine theta and then we have negative g as the vertical acceleration and we have time so let's solve for time we'll have time is equals to if we sort of rearrange this equation it's going to be v's i.

01:28 Sine theta over g.

01:32 And now, since we're obviously multiplying by two, we have that the total time, it's just going to be equal to 2 v.

01:42 I.

01:42 Sine theta over g...

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