# Kinematics: Projectile Motion, Constant Acceleration

• SammyLP250
In summary: So now you can solve for v0y.I solved for v0y by setting the vy equal to zero and the y = 3 in the third Y equation.0 = v0y2 +2(-9.81)(3)v0y = 7.67 m/sI think I may be over thinking this problem because it looks very simple. In summary, the minimum initial velocity and corresponding angle needed to kick the ball just over the 3m high fence is 9.76 m/s and 58.3°.
SammyLP250

## Homework Statement

Determine the minimum initial velocity, v0, and corresponding angle, θ0 needed to kick the ball just over the 3m high fence.

x = 6m
y = 3m

## Homework Equations

X:
vx=v0x
x = x0+v0xt
v0x = v0cosθ
Y:
vy2=viy2
y = y0+v0yt + .5gt2
vy2 = v0y2+ 2g(s-s0)
v0y = v0sinθ

## The Attempt at a Solution

I first started solving for v0y by setting the vy equal to zero and the y = 3 in the third Y equation.

0 = v0y2 +2(-9.81)(3)
v0y = 7.67 m/s

This is where I get stuck at because when I try to go back and plug in v0 = v0y into the third x equation, I'm still left with two variables, θ and v0. I don't know any other way to approach this formula. I searched online and came across a similar problem that used two different formulas I've never seen before:

y = [(v0)sin2θ]/2g

x = [(v0)sin(2θ)]/g

That person said to divide the first formula by the second and solve for angle first, then initial velocity. When I tried that, my answers where θ = 63.4° and v0y = 8.58 m/s.

The book says the correct answers are θ = 58.3° and v0 = 9.76 m/s

Last edited:
It looks simple but it's a cut above the average in difficulty.

Start by writing x and y components of velocity and then distance. You should wind up with 2 equations and 3 unknowns.

The minimum initial speed that will get the ball over the fence does not necessarily correspond to having the ball be at maximum height as it passes over the fence.

Did you mean to say that the answer is vo = 9.76 m/s rather than voy = 9.76 m/s?

TSny said:
The minimum initial speed that will get the ball over the fence does not necessarily correspond to having the ball be at maximum height as it passes over the fence.

Did you mean to say that the answer is vo = 9.76 m/s rather than voy = 9.76 m/s?

Why yes i did :tongue: thank you. And I did not know that, I assumed the ball was at max height at 3m

rude man said:
It looks simple but it's a cut above the average in difficulty.

Start by writing x and y components of velocity and then distance. You should wind up with 2 equations and 3 unknowns.

So I get

v0x = v0cosθ

v0y = vosinθ

and for distance

x = v0xt

y = v0yt + .5(9.81)t2

SammyLP250 said:
Why yes i did :tongue: thank you. And I did not know that, I assumed the ball was at max height at 3m

So I get

v0x = v0cosθ

v0y = vosinθ

and for distance

x = v0xt

y = v0yt + .5(9.81)t2

Good so far EXCEPT for the sign in front of the second term. But a suggestion: don't put numbers in until the end. You lose the ability to check for dimensional compatibility in your terms as you go along. These terms get a bit messy as you will see.

So how about x = v0xt
y = v0yt - gt2/2.
I gave you a 'gimmie' by correcting the sign of the 2nd term in y. You do see why, right?

OK, now how about letting y = h and x = d where d = 6m and h = 3m, and letting t = T since this happens at some instant T. Use d and h, not 6 and 3.

rude man said:
Good so far EXCEPT for the sign in front of the second term. But a suggestion: don't put numbers in until the end. You lose the ability to check for dimensional compatibility in your terms as you go along. These terms get a bit messy as you will see.

So how about x = v0xt
y = v0yt - gt2/2.
I gave you a 'gimmie' by correcting the sign of the 2nd term in y. You do see why, right?

OK, now how about letting y = h and x = d where d = 6m and h = 3m, and letting t = T since this happens at some instant T. Use d and h, not 6 and 3.

Gotcha. And I understand why g is negative.

So they will become

d = v0xT

h = v0yT - gT2

SammyLP250 said:
Gotcha. And I understand why g is negative.

So they will become

d = v0xT

h = v0yT - gT2

Right.
Now you're faced with 2 equations and 3 unknowns (vx0, vy0 and T). What does that suggest to you?

rude man said:
Right.
Now you're faced with 2 equations and 3 unknowns (vx0, vy0 and T). What does that suggest to you?

I'll set T = d/v0x

and substitute T into the second equation.

h = [(v0yd)/v0x] - g (d/v0x)2

So now I'm left with two unknowns!

SammyLP250 said:
I'll set T = d/v0x

and substitute T into the second equation.

h = [(v0yd)/v0x] - g (d/v0x)2

So now I'm left with two unknowns!

Excellent move! Notice you now have vy0 as a function of vx0.

Next step: what are you trying to minimize?

SammyLP250 said:
Gotcha. And I understand why g is negative.

So they will become

d = v0xT

h = v0yT - gT2

Did you lose a factor of 1/2 in the h equation?

rude man said:
Excellent move! Notice you now have vy0 as a function of vx0.

Next step: what are you trying to minimize?

I'm stuck right here. But I think I am trying to minimize v0x

TSny said:
Did you lose a factor of 1/2 in the h equation?

Ahhh That's right. Thank you again. These small mistakes are going to be the death of me.

h = [(v0yd)/v0x] - [g(d/v0x)2/2]

SammyLP250 said:
I'm stuck right here. But I think I am trying to minimize v0x

You need to minimize vo. So, write your equation in terms of vo and θo instead of in terms of vox and voy.

TSny said:
You need to minimize vo. So, write your equation in terms of vo and θo instead of in terms of vox and voy.

Alright then so h = [(v0sinθ0d)/(v0cosθ0)] - (g(d/v0cosθ)2)/2

which become h = tanθ0d - g(d/v0cosθ0)2/2

SammyLP250 said:
Alright then so h = [(v0sinθ0d)/(v0cosθ0)] - (g(d/v0cosθ)2)/2

which become h = tanθ0d - g(d/v0cosθ0)2/2

Yes, good. If you can solve this for vo as a function of θo, you will have a function for which you want to find the minimum.

I'm sorry I haven't replied back. It turns out I wasn't the only one in the class that had trouble with this problem and the teacher realized that and worked it out in class the next day. After looking over it a couple times, I understand what to do now. This problem is definitely not as simple as it looks.

But thank you all for helping me along the way, cause Lord knows I would have just sat there looking at it for hours lol

OK. I hope you enjoy your class.

## What is projectile motion?

Projectile motion is the motion of an object that is launched into the air and then continues to move under the influence of gravity. It follows a curved path known as a parabola.

## What is constant acceleration?

Constant acceleration is when an object's velocity changes by the same amount in a given time interval. In other words, the object's speed increases or decreases by the same amount every second.

## How is projectile motion affected by air resistance?

Air resistance can affect projectile motion by slowing down the object's speed and altering its trajectory. The greater the air resistance, the greater the impact on the object's motion.

## What is the difference between horizontal and vertical components of projectile motion?

The horizontal component of projectile motion is the motion of the object in the horizontal direction, while the vertical component is the motion of the object in the vertical direction. These components are independent of each other, meaning the object's motion in one direction does not affect its motion in the other direction.

## How can the equations of motion be used to analyze projectile motion?

The equations of motion, such as the equations for displacement, velocity, and acceleration, can be used to mathematically analyze projectile motion. By plugging in values for known variables, we can solve for unknown variables and make predictions about the motion of the object.

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