Mechanics in cartesian coordinates

[Ascendant78]

Homework Statement

A cannon shoots a ball at an angle θ above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find the ball's position as a function of time. (Use axes with x measured horizontally and y vertically.) (b) Let r(t) denote the ball's distance from the cannon. What is the largest possible value of θ if r (t) is to increase throughout the ball's flight? [Hint: Using your solution to part (a) you can write down r^2 as x^2 + y^2 , and then find the condition that r^2 is always increasing.]

Homework Equations

x(t) = (v_icosθ)t
y(t) = (v_isinθ)t -1/2gt^2

The Attempt at a Solution

While the part for 'a' was a piece of cake (equations in "relevant equations" above), I am having a hard time figuring out the best way to deal with 'b'. From the hint, I was thinking you take x^2 + y^2, take the derivative of it with respect to θ, then set it to 0 to find what values of θ it will be increasing for. However, I tried this and with how many trig values I ended up with in the equation, I am not even sure how to find the zeros for the function, so I can't discern where it will be zero. Does anyone have any suggestions?

In case it was the right approach, taking the derivative of x^2 + y^2 gave me this:

2v_i^2t^2sinθcosθ - 2v_i^2t^2cosθsinθ - gv_it^3cosθ

 

[Ascendant78]

I guess this is a fairly difficult one, as you guy normally never take this long to toss me some hints. I've skipped over it and done other problems, but still can't figure out what to do on this one.

 

[Tanya Sharma]

Homework Statement

A cannon shoots a ball at an angle θ above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find the ball's position as a function of time. (Use axes with x measured horizontally and y vertically.) (b) Let r(t) denote the ball's distance from the cannon. What is the largest possible value of θ if r (t) is to increase throughout the ball's flight? [Hint: Using your solution to part (a) you can write down r^2 as x^2 + y^2 , and then find the condition that r^2 is always increasing.]

Homework Equations

x(t) = (v_icosθ)t
y(t) = (v_isinθ)t -1/2gt^2

The Attempt at a Solution

While the part for 'a' was a piece of cake (equations in "relevant equations" above), I am having a hard time figuring out the best way to deal with 'b'. From the hint, I was thinking you take x^2 + y^2, take the derivative of it with respect to θ, then set it to 0 to find what values of θ it will be increasing for. However, I tried this and with how many trig values I ended up with in the equation, I am not even sure how to find the zeros for the function, so I can't discern where it will be zero. Does anyone have any suggestions?

In case it was the right approach, taking the derivative of x^2 + y^2 gave me this:

2v_i^2t^2sinθcosθ - 2v_i^2t^2cosθsinθ - gv_it^3cosθ

r² is a function of 't' , not θ . You need to differentiate w.r.t 't' .

 

[ehild]

Homework Statement

A cannon shoots a ball at an angle θ above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find the ball's position as a function of time. (Use axes with x measured horizontally and y vertically.) (b) Let r(t) denote the ball's distance from the cannon. What is the largest possible value of θ if r (t) is to increase throughout the ball's flight? [Hint: Using your solution to part (a) you can write down r^2 as x^2 + y^2 , and then find the condition that r^2 is always increasing.]

Homework Equations

x(t) = (v_icosθ)t
y(t) = (v_isinθ)t -1/2gt^2

The Attempt at a Solution

While the part for 'a' was a piece of cake (equations in "relevant equations" above), I am having a hard time figuring out the best way to deal with 'b'. From the hint, I was thinking you take x^2 + y^2, take the derivative of it with respect to θ

The distance between the ball and the cannon, $R=\sqrt{x^2+y^2}$ has to increase after the ball is shoot out, but it can reach a maximum and then decrease for a while. To have a maximum, dR/dT has to be equal to zero.

Take the derivative of R with respect to time and see if it can be zero.

As R can not be negative, the condition for dR/dT=0 is the same as for d(R²)/dT=0
Apply the chain rule of differentiation...
The first figure shows R(t) and y(t) if V₀=20 m/s and θ=80 degrees. The region where R decreases is clearly seen. The second picture is the same for θ=45 degrees: R inceases during the whole flight.

ehild

ehild

 

[HalfLostAndSearching]

= 0

Simplifying, we get:

2vi^2t^2sinθcosθ - 2vi^2t^2cosθsinθ - gvit^3cosθ = 0

Using the trig identity sinθcosθ = 1/2sin(2θ), we get:

2vi^2t^2(1/2sin(2θ)) - 2vi^2t^2(1/2sin(2θ)) - gvit^3cosθ = 0

This simplifies to:

gvit^3cosθ = 0

We can see that this equation will equal 0 when cosθ = 0, which occurs at θ = π/2. This means that the largest possible value of θ for r(t) to increase throughout the ball's flight is π/2, or 90 degrees. This makes sense, as shooting the ball straight up (90 degrees) will result in the greatest increase in distance traveled.