What is the time of flight for a projectile launched from a cliff?

In summary, the stone is projected from point O with a speed of 20ms^-1 at an angle of elevation of 30 degrees. After T seconds, the stone reaches a depression angle of 45 degrees from O. Using the equation s = ut + 0.5at^2, with s = 0, a = -9.8, and u = 20sin30, we can solve for T and get a value of 2.04 seconds. However, this answer may be incorrect as the angle of depression may actually be -45 degrees and the equation would be \tan \theta = \frac{v_{0y}-gt}{v_{0x}}.
  • #1
brandon26
107
0
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.
 
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  • #2
You've assumption of the angle is wrong.
[tex]\tan \theta = \frac{v_y}{v_x}[/tex]
 
  • #3
Päällikkö said:
You've assumption of the angle is wrong.
[tex]\tan \theta = \frac{v_y}{v_x}[/tex]

Could you expand on that please?
 
  • #5
But how can I show that angle of depression is 45 degrees on that diagram?
 
  • #6
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:
  • #7
Sorry I am confused! Could you go through the question please?
 
  • #8
Positive/negative directions:
+y
|
|
----+x

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


[tex]\tan \theta = \frac{v_{0y}-gt}{v_{0x}}[/tex]
 

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