Projectiles

[brandon26]

A stone is projected from a point O on a cliff with a speed of of 20ms^-1 at an angle of elevation of 30. T seconds later the angle of depression of the stone from O is 45. Find the value of T.

I assumed that when the stone is at depression of 45 degree to O, it has reached the horizontal plane again, hence displacement is 0. so I used s = ut + 0.5at^2.
s=0
a=-9.8
u=20sin30
therefore t= 2.04, which is apparabtly wrong.

Please help.

[Päällikkö]

You've assumption of the angle is wrong.
\tan \theta = \frac{v_y}{v_x}

[brandon26]

You've assumption of the angle is wrong.
\tan \theta = \frac{v_y}{v_x}

Could you expand on that please?

[Päällikkö]

http://img29.imageshack.us/my.php?image=img0071wo.jpg
That's the situation at some point during the flight.

[brandon26]

But how can I show that angle of depression is 45 degrees on that diagram?

[Päällikkö]

You'll want theta = -45o (depending on the positive/negative directions you've chosen).

On the other hand, as tan -45o = -1, vx = -vy (at the desired instant).

[brandon26]

Sorry Im confused! Could you go through the question please?

[Päällikkö]

Positive/negative directions:
+y
|
|
----+x

\tan \theta = \frac{v_y}{v_x}
We also know that
v = v_0+at


\tan \theta = \frac{v_{0y}-gt}{v_{0x}}