Angles of projectiles

[illuminati]

Dear VN Spesh Community,

There's a question I really hate doing; finding the angle a projectile makes with a ground upon collision with the ground (and conversely the angle of direction travelled by a vector).
Does someone know a sure-fire method of doing these?

E.g.

v(t) = 8t i + 10t j - (9.8t+20) k
r(t) = (4t^2 + 1)i + (5t^2) j - (4.9t^2 + 20t) k

Find the angle in which the particle collides with the ground at t = 20/4.9

[golden]

+1

[Vincezor]

Dear VN Spesh Community,

There's a question I really hate doing; finding the angle a projectile makes with a ground upon collision with the ground (and conversely the angle of direction travelled by a vector).
Does someone know a sure-fire method of doing these?

E.g.

v(t) = 8t i + 10t j - (9.8t+20) k
r(t) = (4t^2 + 1)i + (5t^2) j - (4.9t^2 + 20t) k

Find the angle in which the particle collides with the ground at t = 20/4.9

Well first of all, you have to use the velocity vector as it is this that determines the direction the projectile is moving

To find angle of collision (or angle at anytime really) you use the k component if it is 3D and j component if it is 2D and use dot product with the horizontal components (i+j for 3D, i for2D)

Or you can use invtan(y/x) to find the angle at a particular point in time...

[costa94]

velocity vector... sub in time when it hits the ground... you know horizontal and vertical components then... if you still cant figure it out then you lose

[golden]

Dear VN Spesh Community,

There's a question I really hate doing; finding the angle a projectile makes with a ground upon collision with the ground (and conversely the angle of direction travelled by a vector).
Does someone know a sure-fire method of doing these?

E.g.

v(t) = 8t i + 10t j - (9.8t+20) k
r(t) = (4t^2 + 1)i + (5t^2) j - (4.9t^2 + 20t) k

Find the angle in which the particle collides with the ground at t = 20/4.9

Do you have the solutions?

[nubs]

is it 49 degrees?

[illuminati]

oh, nah
I just made the question up on the spot
roflmao
I'll try work it out myself

EDIT: I have 48.93 degrees as an answer

[nubs]

sorry, I forgot to subtract that from 180
I get 131 degrees
This is what I did, a lot of it came down to visualisation
I subbed in the required time into the velocity
I then found the magnitude of the i and j components, which was around 52
I did that^ because we want to find the angle the k component makes with the horizontal axis
So now I have a horizontal line the is 52 units long, and a vertical line that goes 60 units downwards (-60)
So to find that angle, it would be arctan(-60/52)=-49=131 degrees
It's really hard for me to explain, just try and visualise the 3D plane and velocity vector at that particular time, that really helps
Sorry I couldn't explain it any better
Also, I'm not entirely sure I am correct as well so yeah..

[nubs]

If I'm wrong or if there is a better way of going about it, please let me know!

[illuminati]

umm, should still be 49
don't think you can get an obtuse angle for collision with ground if your velocity vector is positive for i, j

[golden]

oh, nah
I just made the question up on the spot
roflmao
I'll try work it out myself

EDIT: I have 48.93 degrees as an answer

I got a similar answer.

[Dangster]

I got 48.94 degrees.
Thanks for this question! I need a lot of practice with projectile motion questions.

[RobM8]

displacement vector's j component isn't 0 meaning the object isn't hitting the ground
it is ~83.30 units above the ground

assuming that j is the vertical component and j component = 0 is the ground

hits ground when 5t^2 = 0 --> t=0 only

[RobM8]

If anyone can be bothered try this one I just conjured:

A particle is launched from the ground at time t=0 from the origin (0,0,0)
Find the acute angle that the particle makes with the ground when it hits it again given that the displacement vector (r) of the particle at time t is given by:

r(t) = (2sqrt(2)t - sqrt(200)t)i + (t^2 - 10t)j + 73tk

Answer: 45 degrees

[yawho]

If anyone can be bothered try this one I just conjured:

A particle is launched from the ground at time t=0 from the origin (0,0,0)
Find the acute angle that the particle makes with the ground when it hits it again given that the displacement vector (r) of the particle at time t is given by:

r(t) = (2sqrt(2)t - sqrt(200)t)i + (t^2 - 10t)j + 73tk

Answer: 45 degrees

If k is the vertical component, the particle will keep on ascending.