Projectile angles and speed (x and Y) near impact

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coneheadceo
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Thought I was done but not the case...

Homework Statement


Alright so as stated a couple days ago... "projectile is shot from the edge of a cliff 125 m above the ground @ an initial speed of 65 m/s at an angle of 37° with the horizontal from the cliff."
---- As helped earlier I have calculated the time it takes for the projectile to hit the ground (10.4 seconds) and on my own the distance it will travel (540 M). But there is more and I am stuck...

" at the moment before impact with the ground find; 1)the horizontal and vertical components of it's velocity, 2)the magnitude of the velocity and 3) the angle made by the velocity vector with the horizontal."


Homework Equations



Not quite sure here

The Attempt at a Solution



I have figured out the Vxf by

Vxo= (65 m/s)(cos 37°)= 51.9 m/s which I equate to the velocity in the x direction for the whole flight so it can apply at impact as well as at launch.

But since the Vyf is in the downward direction I am lost as how to proceed from here to finish the rest of the problem.
 
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To find Vfy you need to use one of the kinematic equations for constant acceleration and apply it on the vertical component of velocity.
 
So would I go something like this?

Vyf = Vyo + at

Vyf = (65 m/s)(sin 37) + (-9.8 m/s^2)(10.4s)

= 39.1 m/s - 101.92 m/s

= -62.82 m/s

answer in the book is -63.1 m/s so I am comfortable with that.

So the overall magnitude of the velocity I would go...

√[(51.9)^2 + (-62.8)^2
= 81.5 m/s

for the overall angle on that beast would I then use...

sin θ = 51.9/ -62.8
= -55.7°

Book says 50.6 below, which is understood because it is negative but I am off with something here.. what am I missing?
 
coneheadceo said:
sin θ = 51.9/ -62.8

Take a second look at this.
 
Flip it?

so go...

tan θ = -62.8/51.9

= -50.4°

Think that will jive... thank you for the help Villyer!
 
You're welcome! And it wasn't just flipping it, you also wrote sin the first time:b