Question

Help, Ive been struggling for a few days with this question!

Answer 1

Suppose a ball moves horizontally at a constant velocity v. Meanwhile, a second ball moves with constant acceleration a along a line oriented at an angle theta from the vertical. At time 0, the first ball is a distance h directly above the second ball, and the second ball is instantaneously at rest.
What angle theta would result in a collision at some later time t? The answer should involve the known values v, h, and a (Let us assume that v, h, and a are all positive).
This collision implies that the positions (x and y coordinates) of both balls are the same at time t. This gives us two equations, one for the x coordinate and one for the y coordinate:
(1/2)at2sin(theta) = v*t
h = (1/2)at2cos(theta)

Answer 2

I dont even know where to start.

Answer 3

your equations look good to me

Answer 4

@Vincent-Lyon.Fr

Answer 5

I'm suppose to solve for theta in terms of v,h,a and t

Answer 6

@ganeshie8

Answer 7

(1/2)at2sin(theta) = v*t
h = (1/2)at2cos(theta)

Answer 8

(1/2)at2sin(theta) = v*t
(1/2)at2cos(theta) = h

divide both equations

Answer 9

tan(theta) = v*t/h

yes ?

Answer 10

Yes

Answer 11

just take tan^(-1) both sides and you're done

Answer 12

Its, not suppose to be in terms of t as I previously stated. sorry

Answer 13

Theta has to be defined by only v,h and a

Answer 14

Ohk.. then you have to solve theta and t

Answer 15

tan(theta) = v*t/h

tan(theta)*h/v = t

Answer 16

substitute this in any of your equations and solve theta

Answer 17

substitute in first equation maybe :

(1/2)at2sin(theta) = v*t

(1/2)a t sin(theta) = v

(1/2)a tan(theta)*h/v sin(\theta) = v

Answer 18

tan(theta)sin(theta) = 2v^2/(a*h)

Answer 19

Okay, but then How would i isolate theta with two trig func
tions in my equation

Answer 20

not sure

Answer 21

hmm okay, well thank you for your help

Answer 22

http://www.wolframalpha.com/input/?i=isolate+%5Ctheta%2C+tan%28%5Ctheta%29sin%28%5Ctheta%29+%3D++2v%5E2%2F%28a*h%29

Answer 23

I find:

\(\tan^2\dfrac{\alpha}{2}=\sqrt{1+(\dfrac{ah}{v^2})^2}-\dfrac{ah}{v^2}\)

Answer 24

The substitution I used, which is very effective in this kind of situation is:
\(u=\tan \dfrac{\alpha}{2}\)

then you can write:

\(\sin \alpha=\dfrac{2u}{1+u^2}\) ; \(\cos \alpha=\dfrac{1-u^2}{1+u^2}\) ; \(\tan \alpha=\dfrac{2u}{1-u^2}\)

You end up with a polynomial that you can love for \(u\):

\(u^4+\dfrac {2ah}{v^2}u^2-1=0\)