Projectile motion on an inclined plane

[steph35]

Homework Statement

A projectile is fired up an incline (incline angle ϕ) with an initial speed vi at an angle θi with respect to the horizontal.

Homework Equations

For what value of θi is d a maximum, and what is that maximum value? (Use v_i for vi, phi for ϕ, and g as appropriate. ϕ is in radians.)

θ>">i =
dmax =

The Attempt at a Solution

Ok so I get so confused with these problems...I did it but the program i have to put it into keep saying syntax error, I don't know if i am wrong or if the program just isn't reading it right...

 

[alphysicist]

Hi steph35,

Homework Statement

A projectile is fired up an incline (incline angle ϕ) with an initial speed vi at an angle θi with respect to the horizontal.

Homework Equations

For what value of θi is d a maximum, and what is that maximum value? (Use v_i for vi, phi for ϕ, and g as appropriate. ϕ is in radians.)

θ>">i =
dmax =

The Attempt at a Solution

Ok so I get so confused with these problems...I did it but the program i have to put it into keep saying syntax error, I don't know if i am wrong or if the program just isn't reading it right...

If it is responding with syntax error, I would think that you are just not inputting your answer in the form that the program needs it to be in.

 

[baba89]

I'm so lost!

I would suggest breaking down the problem into smaller parts and using the appropriate equations to solve each part. For projectile motion on an inclined plane, we can use the equations for projectile motion and the components of motion on an inclined plane.

First, we can find the initial velocity components in the x and y directions:
vx = vi*cos(θi)
vy = vi*sin(θi)

Next, we can use the equations for projectile motion to find the time of flight and the horizontal distance traveled:
t = 2*vy/g
d = vx*t = (vi*cos(θi))*(2*vy/g)

To find the maximum distance, we can take the derivative of d with respect to θi and set it equal to 0:
d' = (2*vi*sin(θi)*cos(θi)^2)/g - (vi*cos(θi)^3)/g = 0
Simplifying, we get:
tan(θi) = 2*tan(ϕ)
Solving for θi, we get:
θi = arctan(2*tan(ϕ))

Plugging this value into the equation for d, we get the maximum distance:
dmax = (vi^2*sin(2*arctan(2*tan(ϕ))))/g