Kinematics: Projectile Motion, Constant Acceleration

[SammyLP250]

Please help. I think I may be over thinking this problem because it looks very simple.

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:
v_x=v_0x
x = x₀+v_0xt
v_0x = v₀cosθ
Y:
v_y²=v_iy²
y = y₀+v_0yt + .5gt²
v_y² = v_0y²+ 2g(s-s₀)
v_0y = v₀sinθ

The Attempt at a Solution

I first started solving for v_0y by setting the v_y equal to zero and the y = 3 in the third Y equation.

0 = v_0y² +2(-9.81)(3)
v_0y = 7.67 m/s

This is where I get stuck at because when I try to go back and plug in v₀ = v_0y 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 v_0y = 8.58 m/s.

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

 

[rude man]

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.

 

[TSny]

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 v_o = 9.76 m/s rather than v_oy = 9.76 m/s?

 

[SammyLP250]

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 v_o = 9.76 m/s rather than v_oy = 9.76 m/s?

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

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

v_0x = v₀cosθ

v_0y = v_osinθ

and for distance

x = v_0xt

y = v_0yt + .5(9.81)t²

 

[rude man]

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



So I get

v_0x = v₀cosθ

v_0y = v_osinθ

and for distance

x = v_0xt

y = v_0yt + .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 = v_0xt
y = v_0yt - 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.

 

[SammyLP250]

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 = v_0xt
y = v_0yt - 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 = v_0xT

h = v_0yT - gT²

 

[rude man]

Gotcha. And I understand why g is negative.

So they will become

d = v_0xT

h = v_0yT - gT²

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

 

[SammyLP250]

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/v_0x

and substitute T into the second equation.

h = [(v_0yd)/v_0x] - g (d/v_0x)²

So now I'm left with two unknowns!

 

[rude man]

I'll set T = d/v_0x

and substitute T into the second equation.

h = [(v_0yd)/v_0x] - g (d/v_0x)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?

 

[TSny]

Gotcha. And I understand why g is negative.

So they will become

d = v_0xT

h = v_0yT - gT²

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

 

[SammyLP250]

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 v_0x

 

[SammyLP250]

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 = [(v_0yd)/v_0x] - [g(d/v_0x)²/2]

 

[TSny]

I'm stuck right here. But I think I am trying to minimize v_0x

You need to minimize v_o. So, write your equation in terms of v_o and θ_o instead of in terms of v_ox and v_oy.

 

[SammyLP250]

You need to minimize v_o. So, write your equation in terms of v_o and θ_o instead of in terms of v_ox and v_oy.

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

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

 

[TSny]

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 v_o as a function of θ_o, you will have a function for which you want to find the minimum.

 

[SammyLP250]

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

 

[TSny]

OK. I hope you enjoy your class.