How Does Projectile Motion Affect a Golf Ball's Landing on a Sloped Terrain?

[vande060]

A golfer drives a ball horizontally with
initial velocity v = (50m/s , 0) from a tee at the
origin, down a 20deg below-horizontal slope as
illustrated above.

A. How far from the tee measured along
the slope does the ball land on the slope?
B. With what speed does it land?

a link to the page in the book with the problem: http://s861.photobucket.com/albums/ab174/alkaline262/?action=view&current=chapter2question.jpg
x(t) = x0 + v0x*t + 1/2*a*t^2
y(t) = y0 + v0y*t + 1/2*a*t^2

x(t) = v0*t
y(t) = -1/2*g*t^2

solving x(t) for t:

t = (x/v0)

inserting in y equation fro trajectory:

y(x) = -1/2*g*(x/v0)^2

i will call the projectile range R, as it would be along the x axis. to compensate for the slope of the hill i will now substitute into the trajectory equation:

R*sin(θ) = -1/2*g*([R^2*cos(θ)^2]/v0^2)

solving for R i get (sin(θ)*-2*v0^2*)/(cos^2*g) = R

inserting v0=50 and θ=-20deg i get the range to be 197

my prof says the range is 185, which i would have gotten i the cos of the denominator was not squared, where is my error here?

I am still working on the second part, but need to clear this up first. Any help would be greatly appreciated

 

[kuruman]

y(x) = -1/2*g*(x/v0)^2

Consider the above equation. You also know that tan20^o = -y/x. So ...

 

[vande060]

Consider the above equation. You also know that tan20^o = -y/x. So ...

xtan(-20) = -1/2*g(x^2/v0^2)

x = [2*v0^2*tan(-20)]/-g = 185

to find the intersection, when the y vales are equal. this comes out to 185, but i don't understand if my math above(when i did it the long way) is incorrect. how did i make a mistake? was it in the math, or was that just altogether incorrect? Thanks for your help so far.

 

[kuruman]

... but i don't understand if my math above(when i did it the long way) is incorrect. how did i make a mistake? was it in the math, or was that just altogether incorrect? Thanks for your help so far.

I am not sure where you got this

R*sin(θ) = -1/2*g*([R^2*cos(θ)^2]/v0^2)

That's probably where you went wrong.

 

[vande060]

I am not sure where you got this

R*sin(θ) = -1/2*g*([R^2*cos(θ)^2]/v0^2)

That's probably where you went wrong.

all right, ill just forget it then. thanks for your help; it was substantial

 

[vande060]

so far we have said

x= [2*v0^2*tan(20)]/g

since tan = -y/x

then y = [-tan(20)^2*2*v0^2]/g

right?

if that is correct, then can i differentiate these two to get the velocities (vx,vy) then plug that into the equation for finding speed of impact :

IvI = √ (vx^2 + vy^2)

if that is right, then it is completely solved!

 

[kuruman]

so far we have said

x= [2*v0^2*tan(20)]/g

since tan = -y/x

then y = [-tan(20)^2*2*v0^2]/g

right?

Right. But x in the above equation is not a function. It is the range R, a constant. Same for y, it stands for the vertical drop when the horizontal distance is R.

if that is correct, then can i differentiate these two to get the velocities (vx,vy) then plug that into the equation for finding speed of impact :

IvI = √ (vx^2 + vy^2)

if that is right, then it is completely solved!

Differentiate what two and with respect to what? The horizontal velocity vx is constant and equal to 50 m/s. You can find the vertical velocity vy if you first find the time of flight and then use vy = gt. Having the two components, you can find the impact angle.

 

[vande060]

welp, back to the drawing board.

185*cos(-20)=v0*t

[185*cos(-20)]/50 = t

3.5s = t

now i believe i can solve for the speed of impact.

given by the equation √(vx^2 + vy^2)

speed of impact = 60.6 m/s

Hopefully that is it, i see what you mean about the derivatives and constants. Thank you for your help