What Is the Required Launch Angle to Clear a Wall in Projectile Motion?

AI Thread Summary
To determine the launch angle required to clear a wall in projectile motion, the problem involves calculating the vertical and horizontal positions of the ball as functions of time. The vertical position equation is y(t) = V[SUB][0]sin(theta)t - (1/2)gt², while the horizontal position is given by x(t) = V[SUB][0]cos(theta)t. It is emphasized that the gravitational constant g can be used in calculations. The approach suggests first ignoring the wall to find the necessary initial velocity for each angle to reach the catcher, and then determining the angle that just grazes the wall. Understanding these relationships is crucial for solving the problem effectively.
ScullyX51
Messages
35
Reaction score
0

Homework Statement


Suppose a very narrow, very tall wall of height H stands between two children who are trying to play catch. The wall is a distance x[wall] from the thrower, and the catcher is distance X[F] from the thrower. In this problem you will calculate the lunch angle (theta) from the horizontal that the thrower must use to just clear the wall and still land at the catcher. You may neglect the height of the children and the width of the wall.
a)Draw a diagram of the problem. (hint: do not assume that the wall's position is halfway between, or that the ball's peak height occurs at the wall)
b0 What is the vertical position of the ball (as a function of time), expressed in terms of the V[0] and the launch angle (theta)?
c) What is the horizontal position of the ball (as a function of time)?

Homework Equations


(delta)y= V[0]yt + .5agt^2
(delta)x= V[0]xt
Vyi= Vsin(theta)
Vxi=Vcos(theta)


The Attempt at a Solution


On the drawing the diagram part I am very confused. I don't know where to begin drawing, if have go by the given hints!

For part B: (the vertical position of the ball as function of time.)
y(t)= V[0]sin(theta)- gt^2/2
I do not think this is right because the problem specifies that the function should only be in terms of theta and v[0]. I do not know how to get this function without g!

For part C: i got the horizontal position of the ball to ne:
x(t)= V[0]cos(theta)t
 
Physics news on Phys.org
ScullyX51 said:
For part B: (the vertical position of the ball as function of time.)
y(t)= V[0]sin(theta)- gt^2/2
I do not think this is right because the problem specifies that the function should only be in terms of theta and v[0]. I do not know how to get this function without g!

For part C: i got the horizontal position of the ball to ne:
x(t)= V[0]cos(theta)t


Hi ScullyX51! :smile:

(have a theta: θ and a squared: ² :smile:)

Your answers are fine.

You are always allowed to use g ! :biggrin:

To continue, first ignore the wall and find what v0 must be for each θ so as to land at the catcher.

Then find out which value of θ grazes the wall. :wink:
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top