If the golf ball of Prob. 2 / 116 is launched at time t=0, determine the two times when the radi

If the golf ball of Prob. 2 / 116 is launched at time t=0,...

Key Concepts

Parametric Differentiation

Parametric differentiation is the process of finding derivatives when both x and y coordinates of a curve are expressed as functions of an independent parameter, often time. This technique is crucial for computing velocities, accelerations, and the curvature of a trajectory in problems involving motion, as it allows for the separate handling of components associated with the motion.

Radius of Curvature

The radius of curvature is a measure of how sharply a curve bends at a given point; mathematically, it is defined as the reciprocal of the curvature of the curve. In the context of a trajectory, it represents the radius of the instantaneous circular path that best approximates the curve at a specific point, calculated using the derivatives of the parametric equations.

Projectile Motion

Projectile motion describes the motion of an object thrown into the air, where its path is determined by its initial velocity, the launch angle, and the acceleration due to gravity. It involves analyzing how both horizontal and vertical components interact over time, commonly using parametric equations to detail the object's trajectory.

Transcript

00:01 So here assuming x x x is to be in horizontal direction and positive towards right and y x is to be in vertical direction and positive upward.

00:10 Let us write the equation of equation for the radius of curvature of trajectory using equation radius of curvature of trajectory r will be equals to x -dice square plus y -dus square to the power of three divided by two divided by x -dice y double dash minus y -dice x double -dess this is to be in mode so here x -dess is x -dust -y -dose are velocity in x and y -dirtion x double -dust and y -dell -dust are acceleration in x and y -direction so let us write r is equal to x -dess square plus y -dase square to the power of three divided by two whole divided by x -dus y -douleds so x -dice y double -d -s is nothing but g.

01:11 It is minus g.

01:14 Excellation due to gravity minus y -d -dase is 0.

01:19 Since there is no acceleration in horizontal direction.

01:23 So let us put it here.

01:26 So it will be equals to x -dose square plus y -d -square whole to the power of 3 divided by 2 divided by x -dase g.

01:41 Let this equation be question number one.

01:46 Now we know that the horizontal component of velocity is u cost theta, u cost, and vertical component of velocity is equals to u sine theta minus zt, minus zt.

02:03 So let us substitute these value in equation one.

02:07 From equation one, from equation one, we can write radius of curvature r is equals to u cost theta square plus u sine theta minus gt square square whole to the power of 3 divided by 2 whole divided by u cost theta multiplied by g so let us substitute the values here radius of curvature will be the value of u u is given as the radius of curvature so let's let's put radius of curvature which are even as 1 ,800 feet.

02:58 So it is 1 ,800 fit...

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