Kinematics ball throwing question

[les3002]

hi I am having some trouble with the question below, was hoping maybe someone could talk me through it?

A ball is thrown from a point on an incline of 16 degrees with an initial velocity of 19m/s and at an angle of 49 degrees to the incline. How far up the plane will the ball strike measured from its initial position?

so far i have resolved the velocity into its x and y directions

Ux = 8.03m/s
Uy = 17.22m/s

So - Sy = Uy*t-(0.5gt^2)
Sx = Ux*t

Now i get stuck with the rest, i have to:

Express the height of the incline, Sp in terms of time, t

Sp = ____ xt

At the point of impact, y = Sp, find t:

t = ____ s

Now find Sx and Sp

Sx = ____ m
Sp = ____ m

Therefore, the distance up the incline from the point where the ball was released is:

Sball = _____ m


i would appreciate any help people can offer me with this as i understand its not that difficult but just can't grasp it.

thanks les.

 

[berkeman]

I don't think I'd use time in this problem. Just write equations for the two lines and set them equal to find the impact point. Write an equation for the y=f(x) of the ball as it flies along, and the y=f(x) straight line of the incline.

 

[radou]

Another way: let the time of impact equal $$t_{0}$$. Then, $$y(t_{0}) = v_{0} \sin(65)t_{0} - \frac{1}{2}gt_{0}^2$$. Further on, from the geometry of the incline, you can conclude $$y(t_{0})=x(t_{0})\tan(16) = v_{0}\cos(65)t_{0} \tan(16)$$. Solve for $$t_{0}$$, the rest is trivial.

 

[les3002]

thanks for your input so far mate, the only problem is i have to answer all the parts above, basically do it in the way they have set it out.