Finding the angle of the velocity vector of a projectile at moment of impact

[Corey333]

Homework Statement

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 16.0 m/s. The cliff is h= 41.0 m above a flat horizontal beach
h=41
g=10m/s
vi=0m/s
vf2=vi2+2gh
vi=28.64
h=vi+vf/2xt
41=28.64/2xt
14.32
t=2.86
28.64x28.64, 16x16
Square root of 1076.24=32.806

I don't know how to find the angle, On my homework it requires the speed of which the impact lands which is 32.806, and the time it takes to strike the cliff after the stone is thrown which is 2.86, but it is asking for the angle below the horizontal but I don't know how to get it PLEASE HELP A.S.A.P.
I have a 90% on my homework right now but I need to find the angle to get a 100%, I'm a gpa freak I really need this help thank-you in advance!

 

[voko]

I assume you mean the angle the velocity vector makes with the horizontal. Find the horizontal and vertical components of velocity, and, given that, find the angle.

 

[Corey333]

I have that, 28.64 and 16, I just don't know what to do with that.

 

[voko]

Draw a triangle, two sides of which are the components of the velocity vector. The hypotenuse in the triangle is the velocity. How are the angles and the sides in the triangle related?

 

[HalfLostAndSearching]

I appreciate your dedication to achieving a perfect score on your homework. To find the angle of the velocity vector at the moment of impact, we need to use the equation for projectile motion: vf = vi + at. In this case, we can use the known values of vf (final velocity of 32.806 m/s), vi (initial velocity of 16 m/s), and t (time of 2.86 seconds). Rearranging the equation, we get a = (vf - vi)/t, which gives us an acceleration of 5.28 m/s^2.

Next, we can use the fact that the horizontal component of the velocity (vx) remains constant throughout the motion and the vertical component (vy) is affected by gravity. We can use the equation vx = v cos(theta) and vy = v sin(theta), where v is the magnitude of the velocity and theta is the angle below the horizontal. We know the magnitude of the velocity (32.806 m/s) and the time (2.86 seconds), so we can solve for the angle by substituting these values into the equations.

vx = 32.806 cos(theta)
16 = 32.806 cos(theta)
cos(theta) = 16/32.806
theta = cos^-1(16/32.806)
theta = 61.9 degrees below the horizontal.

Therefore, the angle of the velocity vector at the moment of impact is 61.9 degrees below the horizontal. I hope this helps you achieve your desired score on your homework. Remember, it's important to understand the concepts and not just aim for a perfect score. Good luck!