Launch angle and Projectiles

[ScullyX51]

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<sub>[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</sub>

 

[tiny-tim]

For part B: (the vertical position of the ball as function of time.)
y(t)= V<sub>[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</sub>

Hi ScullyX51!

(have a theta: θ and a squared: ² )

Your answers are fine.

You are always allowed to use g !

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. ​

 

[baba89]

This is correct, as it only depends on the initial velocity and launch angle, not the acceleration due to gravity.

I would like to clarify a few things about launch angle and projectiles.

Firstly, the launch angle is the angle at which an object is projected into the air. It is measured from the horizontal and can greatly affect the trajectory of the object.

In this particular problem, the launch angle must be calculated in order for the ball to clear the wall and reach the catcher. This can be done by using the equations for projectile motion, as shown in the attempt at a solution.

However, it is important to note that in real life situations, the height and width of the wall, as well as air resistance, would also affect the trajectory of the object and must be taken into account. This problem assumes ideal conditions, but in reality, these factors would need to be considered to accurately determine the launch angle.

Furthermore, the launch angle is not the only factor that affects the trajectory of a projectile. The initial velocity, air resistance, and the mass of the object also play a role.

In conclusion, while the launch angle is an important factor in determining the trajectory of a projectile, it is not the only factor and must be considered in conjunction with other factors to accurately predict the motion of an object.