# Projectile Motion: Finding the Best Angle for a Tennis Ball Launcher

• db2dz
In summary, a tennis ball launcher with a speed of 35 m/s is used to try and hit a window 10 m below the current height of a building that is 100 m away. The time it takes for the ball to reach the building is 2.86 seconds. The ball does not go through the window, instead it hits the building 30 m below the window. To aim the launcher and make the ball go through the window, two angles can be used: one angle is 19.2° and the other angle is greater than 19.2°. The time for the ball to reach the window at 19.2° is 3.03 seconds. The other time is longer than 3
db2dz

## Homework Statement

You are on the roof of a building that is 150 m tall armed with a tennis ball launcher that fires tennis balls with a speed of 35 m/s. Another building is 100 m away from your building. You are trying to fire a tennis ball into a window of the other building that is 10 m below your current height. Knowing that the ball will fall as it travels, you first align the launcher so that it is completely horizontal, hoping that it will fall the necessary 10 m as it reaches the building.

(a) How long does it take for the ball to reach the building?
(b) Does the ball go through the window? If not, where does the ball hit the building?
(c) If you want the ball to go through the window, at what angles could you aim the launcher? (NOTE: There are two angles that work! Find both!)
(d) For each angle in (c), find the time for the ball to reach the window.
(e) Sketch the two trajectories that the ball could take to go through the window. Explain why the times you found in (d) are different.

ANSWERS: (a) 2.86 s (b) no, 30 m below window (c) One angle is 19.2°. The other is greater than this.
(d) The time for 19.2º is 3.03 s. The other time is longer than this. (e) Show me. Tell me.

## Homework Equations

x=xo + vxo*t
y= yo + vyo* t - (1/2)gt

## The Attempt at a Solution

i know what to do for A and B, but i am on C and i don't know how to find the angle. My teacher gives the answers so i know one of the angles but i don't know how he got it.

Welcome to PF.

You have the 2 equations and you know that t is the same between both.

You have Vo, and you know x-velocity then is Vo*Cosθ and the y component is Vo*Sinθ

The only thing I suggest then is to substitute the values you know (basically everything but θ and t) and grind it out.

You will possibly find these relationships useful ...

2*Cosθ*Sinθ = Sin2θ
Cos²θ = 1/2(1 + Cos2θ)
tanθ = Sinθ/Cosθ

LowlyPion, have you solved this yourself? If so, could you possibly PM me the correct and full answer? I've been trying to crack this myself and it drives me nuts.

Kruum said:
LowlyPion, have you solved this yourself? If so, could you possibly PM me the correct and full answer? I've been trying to crack this myself and it drives me nuts.

Sorry, solutions to homework in private mail are against the spirit - not to mention the rules - of the forum.

I provided those trig identities as possibly useful to get to the final answer.

Perhaps you will find them of value as well?

LowlyPion said:
Sorry, solutions to homework in private mail are against the spirit - not to mention the rules - of the forum.

This is not my homework, I'm trying to solve it just for fun. And those trig identities you gave was nothing new to me. The only problem I'm having with solving this is that I have sin(2x) and cos(2x) left and I can't get them so that I'm only left with tan(2x).

ok i got this in class.
you have to use y= Yo + x(tanθ)-(.5)(g)(x/Vocosθ)^2
for this y=0
x=100
Vo=35

put it in your graphic cal and find the zeroes

Kruum said:
This is not my homework, I'm trying to solve it just for fun. And those trig identities you gave was nothing new to me. The only problem I'm having with solving this is that I have sin(2x) and cos(2x) left and I can't get them so that I'm only left with tan(2x).

No I didn't solve it, but I seem to recall solving a similar problem using the Sin2x relationship. For instance this is how the usually derived range equation is derived.

I'll take a look at solving it.

db2dz said:
put it in your graphic cal and find the zeroes

Yikes. That's allowed?

OK. Forget the 2θ identities.

There's one additional identity that can be used I see now, and it leads directly to the the solution.

Sin2θ + Cos2θ = 1

That means that when you get to the equation that looks like

-10Cos2θ = 100*SinθCosθ - 40

Substitute in for the -40 = -40*(Sin2θ + Cos2θ)

Then solve the quadratic in Sinθ, Cosθ. That leads to the 2 factors that you can then calculate the angles using tan-1.

Sorry if my misdirection caused any headaches.

you don't really need to use any identities

we know that
x=xo + Vxo(t)

so time = x/Vxo because xo in this problem = 0
and Vxo= VoCosθ
so t= x/VoCosθ

we also know that
y= yo+Vyo(t)-(.5)(g)(t)^2

so plug in t= x/VoCosθ wereever their is time

which would look like
y= yo+Vyo( x/VoCosθ)-(.5)(g)( x/VoCosθ)^2

and Vyo= VoSinθ

so it becomes
y= yo+ (x)(Tanθ)-(.5)(g)( x/VoCosθ)^2

Solve for θ
and you get the angles

This is how my class worked it out, I am sure there is another way for it to work.

db2dz said:
Solve for θ
and you get the angles

This is how my class worked it out, I am sure there is another way for it to work.

OK. Solve for θ.

What are the steps that you used to algebraically reduce the quadratic in tanθ and cosθ.

If your solution is to plug it into a calculator, then I quite think you've missed the point of the substitution identities.

My solution yields both of the angles directly ... and without a scientific calculator.

well I am not sure how algebraically reduce the quadratic off the top of my head and i haven't missed the point of the substitution identities. the way my class used involves substiutions but with the equations and knowns to make a new equations. And i think it is cool that you found way to do it without a calculator, but you get the same results using the calculator and my teacher doesn't mind us using the calculator.

db2dz said:
y= yo+ (x)(Tanθ)-(.5)(g)( x/VoCosθ)^2

Solve for θ
and you get the angles

I would like to see how to solve this one. My school forbids using any sorts of calculators in exams, so I really need to practice without a calculator.

Kruum said:
I would like to see how to solve this one. My school forbids using any sorts of calculators in exams, so I really need to practice without a calculator.

This earlier post provides the solution directly.

https://www.physicsforums.com/showpost.php?p=2075099&postcount=9

Plugging the substitution yields

30*cos²θ -100*Sinθ*cosθ + 40*Sin²θ = 0

This yields 2 solutions to the quadratic, that which when set to 0, provides results of the form

tanθ = k

Taking the tan-1(k) = ∠

LowlyPion said:
This earlier post provides the solution directly.

Oh, sorry. I seem to have missed that post. Thank you.

## 1. What is 2-D projectile motion?

2-D projectile motion refers to the motion of an object that is launched or thrown at an angle to the ground, and moves along a curved path due to the influence of gravity.

## 2. What factors affect 2-D projectile motion?

The factors that affect 2-D projectile motion include the initial velocity of the object, the angle at which it is launched, the acceleration due to gravity, and the air resistance acting on the object.

## 3. How is the range of a projectile calculated?

The range of a projectile can be calculated using the formula R = (v^2 * sin(2θ))/g, where R is the range, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

## 4. What is the maximum height reached by a projectile?

The maximum height reached by a projectile can be calculated using the formula h = (v^2 * sin^2(θ))/2g, where h is the maximum height, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

## 5. How does air resistance affect 2-D projectile motion?

Air resistance can affect 2-D projectile motion by slowing down the horizontal velocity of the object and reducing the overall range. It also causes the object to follow a slightly different trajectory compared to when there is no air resistance.

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