# Projectile Motion

1. Sep 17, 2010

### XwakeriderX

1. The problem statement, all variables and given/known data

See picture

2. Relevant equations
Trajectory equation and total time of flight

3. The attempt at a solution
Okay what would be the best way to approach this? should i find out the distance the ball goes? then subtract 35m to to get my length of hill and use that for cos30=x/h

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2. Sep 17, 2010

### XwakeriderX

Please anyone ive been at this for hours

3. Sep 17, 2010

### PhanthomJay

Which trajectory equations do you think apply? Note: I see from your prior posts that the problems you are submitting are of a somewhat higher level of difficulty than introductory level basic problems. What level of physics are you taking?

4. Sep 17, 2010

### XwakeriderX

Just a normal college physics class "Mechanics-Solids/Fluids" at a junior college. My teacher does not give any examples or anything. This is his old test questions so im trying to figure them out because they are not like normal two dimension or trajectory problems! Are you not watching the game!? i have to study for physics my test is on wednesday! Okay now back on topic

I know the basic trajectory equations

Y=Yo + Xtan(THETAo)- (gx^2/(2Vo^2Cos^2THETAo)

I think i want to see where the golf ball is at the base of the hill an then add the rising mountain to the downward acceleration

5. Sep 17, 2010

### PhanthomJay

I would be watching the game, but I'm trying to assist you! But in the meantime , I have to run some errands, so rather than leave you hanging, I'll see if someone else can assist. Hang in there.

6. Sep 17, 2010

### XwakeriderX

thanks! i really appreciate it! I'm going premed but UCLA wants me to have physics under by belt which doesnt really make sense unless i want to find out how far a kid fell of his bike to break his arm...haha

7. Sep 17, 2010

### Staff: Mentor

What game?

One key hint on trajectory questions like this is to see that the horizontal velocity remains constant when there is no air resistance. The vertical velocity does vary, according to the constant acceleration kinematic equations. So the ball follows a parabolic arc. You need to find the equation for that arc, and equate it with the equation of the line for the hill's slope. Where the two lines meet, that's where the ball hits.

See "Kinematics of constant acceleration" on this page:

http://en.wikipedia.org/wiki/Kinematics

.

8. Sep 17, 2010

### XwakeriderX

red sox!

Okay that makes some sense to me, ill try finding the equation to the arc

9. Sep 17, 2010

### XwakeriderX

Range = [sin(2theta)V(initial)^2]/ acceleration ????

10. Sep 17, 2010

### vela

Staff Emeritus
Set up your coordinates so the point (x,y)=(0,0) is at the base of the hill. The hill is then given by the line y=mx for some value of m, which you can figure out. Tell us what the equations for x(t) and y(t) of the ball should be.

11. Sep 17, 2010

### XwakeriderX

d = 35.0 m,
angle A = 45.0 degrees,
angle B = 30.0 degrees,
initial velocity v = 145*0.447 m/s
find h

horizontal displacement x = d + h/tan(B) = vcos(A)*t (1)
vertical displacement y = h = vsin(A)*t - gt^2/2 (2)
from (2),
gt^2/2 - vsin(A)*t + h = 0
solve it and get
t = {vsin(A) + sqrt[v^2*sin(A)^2 - 2gh]}/g
put it into (1), get
d + h/tan(B) = vcos(A)*{vsin(A) + sqrt[v^2*sin(A)^2 - 2gh]}/g
gd + gh/tan(B) = v^2*cos(A)*sin(A) + v*cos(A)*sqrt[v^2*sin(A)^2 - 2gh]}
insert known values, simplify and get
sqrt(2100 - 19.6h) = 0.37037h - 38.337
squared both sides and simplify,
0.13717h^2 - 8.79h - 630.3 = 0
solve it and get
h = 107 m

:)