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

Click For Summary
SUMMARY

The discussion focuses on calculating the required launch angle (θ) for a projectile to clear a wall of height H while landing at a distance X[SUB][F] from the thrower. The equations of motion used include the vertical position function y(t) = V[SUB][0]sin(θ)t - (1/2)gt² and the horizontal position function x(t) = V[SUB][0]cos(θ)t. Participants clarified that gravitational acceleration (g) can be included in calculations, and the approach involves first determining the necessary initial velocity (V[SUB][0]) for various angles before identifying the angle that just grazes the wall.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of trigonometric functions related to angles
  • Basic understanding of gravitational effects on motion
NEXT STEPS
  • Calculate the initial velocity (V[SUB][0]) required for different launch angles (θ) in projectile motion
  • Explore the concept of maximum height in projectile motion and its relation to launch angle
  • Study the effects of varying wall heights (H) on the required launch angle
  • Learn about optimization techniques in projectile motion to minimize launch angle while clearing obstacles
USEFUL FOR

Students studying physics, educators teaching projectile motion concepts, and anyone interested in solving real-world problems involving trajectories and angles in motion.

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:
 

Similar threads

  • · Replies 36 ·
2
Replies
36
Views
4K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
25
Views
3K
Replies
12
Views
1K
  • · Replies 39 ·
2
Replies
39
Views
5K
Replies
10
Views
2K
  • · Replies 15 ·
Replies
15
Views
27K
  • · Replies 16 ·
Replies
16
Views
4K
Replies
7
Views
2K