Find initial min launch velocity with coeffcient of restitution

[vxfuriousxv]

Homework Statement

Hi new to the forum, I am trying to make a ball bounce into a cup by hitting it with a paddle, giving it [V1 @ theta 1 wrt horizontal]. It travels a vertical distance delta y1 and a horizontal distance delta x1 before stricking the table in a semi-elastic collision characterized by a coefficient of restitution [e]. The ball bounces off the table surface and reaches the apex of flight delta y2. just as it crosses the lip of the cup delta x2 from the collision site. Show that the minimum v1 required can be written as sqrt[(2g(delta y2) - (e^2)(delta y1)]/ esin[tan^-1((2/e delta x2)sqrt((delta y2^2)+(e^2)(delta y1)(delta y2))] where theta 1 = tan^1... taken with respect to the horizontal.

Homework Equations

e= (V1f - V2f)/(V2o - V1o)

The Attempt at a Solution

So far i have Vio = sqrt[(V1f^2) - 2g(delta y1)]/sin theta 1. V2f = 0 therefore 0= (V2o^2) - 2g delta y2. Since the problem is semi elastic, then V1f does not equal to V2o.
...and that's it. I am confused on how to incorporate the solution with e. Please help. Thanks.

 

[anathema]

Hi there, welcome to the forum! This is an interesting problem you have here. Let's break it down step by step.

First, let's define our variables:
V1 = initial velocity of the ball before hitting the table
theta 1 = angle of V1 with respect to the horizontal
V1f = final velocity of the ball after hitting the table
V2o = initial velocity of the ball after bouncing off the table (since it is a semi-elastic collision, V1f does not equal V2o)
V2f = final velocity of the ball after reaching the apex of flight
g = acceleration due to gravity
delta y1 = vertical distance traveled by the ball before hitting the table
delta x1 = horizontal distance traveled by the ball before hitting the table
delta y2 = vertical distance from the apex of flight to the lip of the cup
delta x2 = horizontal distance from the collision site to the lip of the cup
e = coefficient of restitution

Next, let's use the equation for coefficient of restitution to relate V1f and V2o:
e = (V1f - V2f)/(V2o - V1o)
Since V1f does not equal V2o, we can rearrange this equation to solve for V2o:
V2o = (V2f - eV1f)/(1-e)

Now, let's consider the motion of the ball after it bounces off the table. We know that the vertical component of its velocity will be equal to V2o, since it is at the apex of flight. The horizontal component will depend on the angle of the ball's trajectory and its initial velocity. We can use the following equations to relate V2o and the angle of the ball's trajectory with respect to the horizontal (theta 2):
V2o = V2f*sin(theta 2)
V1f = V2f*cos(theta 2)

Now, let's consider the horizontal distance traveled by the ball from the collision site to the lip of the cup, which is delta x2. We can use the equation for horizontal distance to relate delta x2, V2f, and theta 2:
delta x2 = V2f*cos(theta 2)*t
where t is the time it takes for the ball to reach the lip of the cup.

We can also use the equations for vertical distance to relate delta y2, V

 

[kerimek]

Hello and welcome to the forum! I would like to help you with your problem.

First, let's define some variables for clarity:
- V1o = initial launch velocity
- V1f = final velocity after the first bounce off the table
- V2o = initial velocity after the first bounce off the table
- V2f = final velocity at the apex of the second bounce
- delta y1 = vertical distance traveled before the first bounce
- delta y2 = vertical distance traveled before reaching the apex of the second bounce
- delta x1 = horizontal distance traveled before the first bounce
- delta x2 = horizontal distance traveled before reaching the lip of the cup
- e = coefficient of restitution

Now, let's start by using the equation for coefficient of restitution:
e = (V1f - V2f)/(V2o - V1o)

We can rearrange this equation to solve for V1f:
V1f = V2f + e(V2o - V1o)

Next, let's consider the motion of the ball after the first bounce. It will travel a horizontal distance of delta x1 and a vertical distance of delta y2 (since it reaches the apex of the second bounce). We can use the equations of motion to relate these distances to the initial and final velocities:
delta x1 = (V1f^2 - V2o^2)/(2g)
delta y2 = (V1f^2 - V2f^2)/(2g)

Substituting the expression for V1f that we found earlier, we get:
delta x1 = ((V2f + e(V2o - V1o))^2 - V2o^2)/(2g)
delta y2 = ((V2f + e(V2o - V1o))^2 - V2f^2)/(2g)

Now, using the given relationship between delta x2 and delta y2, we can write:
delta x2 = (delta y2/e) * tan(theta 1)

Substituting this into the equations for delta x1 and delta y2, we get:
delta x1 = ((V2f + e(V2o - V1o))^2 - V2o^2)/(2g)
delta y2 = ((V2f + e(V2o - V1o))^2 - V2f^2