- #1
merlinMan
- 13
- 0
A ball is thrown from a bliff with an initial speed of 20m/s from the edge of a 45m high cliff. At the same time, a woman is running away from the cliff at a speed of 6m/s. She runs until she catches the ball. at what angle above the horizon should the ball be thrown so that she can catch ball.
I came up with a position function for the woman
[tex] V_{w} = 6t [/tex]
and set it equal to the vertical position function for the ball.
[tex] 6t= 45+ 20sin(X)t -\frac{-9.8}{2}t^2 [/tex]
I can't see where I go from here. I tried using quadratic but I didn't think I could combine the 6t and the 20sin(x)t, am I incorrect? How do I do it with 2 variables, the x and the t?
I came up with a position function for the woman
[tex] V_{w} = 6t [/tex]
and set it equal to the vertical position function for the ball.
[tex] 6t= 45+ 20sin(X)t -\frac{-9.8}{2}t^2 [/tex]
I can't see where I go from here. I tried using quadratic but I didn't think I could combine the 6t and the 20sin(x)t, am I incorrect? How do I do it with 2 variables, the x and the t?