- #1
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 = Vo2 +2a(x-xo)
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.