A rock is thrown upward from level ground in such a way that...

A rock is thrown upward from level ground in such a way that the maximum height of its flight is equal to its horizontal range $R$ (a) At what angle $\theta$ is the rock thrown? (b) In terms of its original range $R,$ what is the range $R_{\max }$ the rock can attain if it is launched at the same speed but at the optimal angle for maximum range? (c) What If? Would your answer to part (a) be different if the rock is thrown with the same speed on a different planet? Explain.

Key Concepts

Optimization of Launch Angle

Optimizing the launch angle involves determining the angle that maximizes a particular aspect of the projectile’s motion, such as horizontal range. This requires setting up relationships between the maximum height and range formulas and often involves finding when one variable (like the sine function of a multiple angle) is maximized. In the absence of additional constraints, the optimal angle for maximum range on level ground is typically 45°.

Maximum Height and Horizontal Range

The maximum height in projectile motion is achieved when the vertical component of the initial velocity is completely counteracted by gravity, while the horizontal range is the total distance traveled horizontally. These are given by formulas that depend on the initial speed, launch angle, and gravitational acceleration, and understanding these formulas is essential to analyze projectile trajectories.

Independence from Gravitational Effects in Angle Determination

In many projectile motion problems, the gravitational acceleration appears in both the height and range formulas and cancels out when finding relationships such as the optimal launch angle. This means that certain results, like the optimum angle determined by a specific condition, can be independent of the value of gravitational acceleration, making the analysis applicable under different gravitational conditions.

Projectile Motion

Projectile motion is the motion experienced by an object that is launched into the air and moves under the influence of gravity alone, with its horizontal and vertical components of motion being independent. This concept is foundational for analyzing trajectories, determining time of flight, and relating the displacement of objects in both horizontal and vertical directions.

Kinematic Equations for Projectile Motion

The kinematic equations describe the motion of projectiles. They relate initial velocity, the angle of launch, acceleration due to gravity, time, and displacement. These equations provide formulas for key aspects such as maximum height and horizontal range, which are critical for solving projectile problems.

Transcript

00:01 Okay, so we're throwing a rock up from the ground.

00:06 It's going to have initial velocity and a sine theta, i mean an angle of theta.

00:15 If we want the range to equal its maximum height, then what would theta be? well, i'm going to start with y, vy at the peak is going to equal vy0, which would be v0 sine theta times t minus one -half gt squared.

00:51 But wait a minute, why do i need to get t into this? vy peak squared is v0 squared.

00:58 So this is another equation for projectile motion.

01:08 V0 squared, v squared equals v.

01:11 Zero squared minus 2gh.

01:19 Okay, so this is zero.

01:27 And therefore, the height is going to be v0 squared, sine squared, sign squared, of theta over 2g, but we want the range to be equal to the height.

01:54 So i'm noticing that i put too many t's in here, and i kind of messed up the equation.

02:02 I do want to go back to this.

02:04 V equals v initial minus gt to the peak.

02:11 So t to reach the peak is going to be v0, sine theta, over g v0 sine theta over g okay but time to reach the whole way time final is going to be twice that much over g okay the range um x is going to be vx times t final vx times t final vx is v0 cosine theta times t final, which is 2v0 sine theta over g.

03:06 That would be the range.

03:09 But the range, we already know we want it to be h.

03:12 So we want that to be v0 squared, sine squared theta, over 2g.

03:21 The gs cancel out, which seems strange.

03:39 Okay.

03:42 A sine theta cancels out from both sides.

03:47 So this is going to give us, actually, a v0 cancels out from both sides.

04:00 Both v zeros cancel out from both sides.

04:09 So, i think this is telling me that the tangent of theta equals four.

04:30 So theta is going to be the inverse tangent of four, which is 76 degrees.

05:01 I want to look at significant digits.

05:03 Well, the only thing we know is we want them to be the same.