Find angle between displacement and velocity vector

[soupleaf]

Homework Statement

A stone is thrown at 25m/s and at 37 degrees above the horizontal from a 20m high building. Take g=10m/s²
Let D be the displacement vector from launch point on top of the building to landing point on the ground, and v be the velocity vector on impact, what is the angle between these two vectors?

I have the answer being 51.34 degrees - 14.04 degrees = 37.30 degrees from my teacher!

I have already found that vertical velocity is 15.05m/s, horizontal velocity is 19.97m/s. It takes 1.5s to reach the high point, 4s to hit the ground, goes 80m horizontal distance before hitting the ground, and its speed of impact is sqrt 1025 or 32.02m/s.

Homework Equations

V² = V_o² +2a(x-x_o)

The Attempt at a Solution

I know that D as a vector can be represented as seen below because of the given building height and the stone horizontal distance.

To get theta I just did arc tan (20/80)

v as a vector I am calculating its components at impact to be...

vertical/j hat is V² = (15.05)² + 2(10)(20)
V = 25.03 m/s j hat

horizontal/i hat is V² = (19.97)² +2(0)(80)
V = 19.97 m/s i hat

So to find the angle of the similar right triangle would be arc tan (25.03/19.97) = 51.41 degrees

Since the 2 triangles overlap because they both connect at the impact site, I do 51.41 degrees - 14.04 degrees to get the distance between v and D.
However this goes against the given answer my teacher gave me of 51.34 degrees - 14.04 degrees = 37.30?

I'm confused because I'm tried thinking of different ways to do it, plugging in any of the values I've solved for or been given that are appropriate, but I can never get 51.34 degrees.

 

[kuruman]

To get theta I just did arc tan (20/80)

That's totally incorrect. Your drawing shows that the projectile follows a triangular path and that it hits the ground at the same level from which it was launched. You need to reconsider. Hint: $\vec{D}$ is a position vector that depends on time. Can you find expressions for $\vec{D}(t)$ and $\vec{v}(t)$?

 

[soupleaf]

That's totally incorrect. Your drawing shows that the projectile follows a triangular path and that it hits the ground at the same level from which it was launched. You need to reconsider. Hint: $\vec{D}$ is a position vector that depends on time. Can you find expressions for $\vec{D}(t)$ and $\vec{v}(t)$?

Thank you very much for the reply. I found out my own error. It said take g=10m/s² but I guess I messed up my coord system somewhere making down +. I went back and made down - so g=-10m/s² and everywhere else in my work and in the end vertical V should have been calculated using V= V_o + at.
V = 15.05 + (-10)(4) = -24.95 m/s.
arc tan (-24.95/19.97) = 51.34 degrees

51.34 - 14.04 = 37.70 degrees which was the answer my teacher gave.

My drawing did not have enough detail maybe because in the end the calculator for theta was correct. It was launched from a building 20m high which I did not include in the drawing, and the theta I calculated was the angle at the bottom near the impact site. Maybe that clears it up, not sure...

 

[haruspex]

It said take g=10m/s2 but I guess I messed up my coord system somewhere making down +. I went back and made down - so g=-10m/s2 and everywhere else

Yes, you had the sign wrong, but that description is backwards. If down is positive then g=+10m/s², but the initial velocity is negative, etc.