# Assistance with Projectiles

1. Oct 1, 2005

### 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.

2. Oct 1, 2005

### Päällikkö

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

3. Oct 1, 2005

### brandon26

Could you expand on that please?

4. Oct 1, 2005

5. Oct 1, 2005

### brandon26

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

6. Oct 1, 2005

### 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).

Last edited: Oct 1, 2005
7. Oct 1, 2005

### brandon26

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

8. Oct 1, 2005

### Päällikkö

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

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

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