A hockey puck slides off the edge of a table with an initial velocity of 26.0 m/s. The height of the table above the ground is 2.00 m. What is the angle below the horizontal of the velocity of the puck just before it hits the ground?

Question

anonymous · Answer

Can someone please help me with this question...

darklight · Answer

I think it's 6.87 degrees below the horizontal and just before it hits the ground it should have a downward speed of 6.2524 meters per second

anonymous · Answer

These are the options 

a. 12.8°

 b. 13.5°

 c. 31.8°

 d. 77.2°

 e. 72.6°

darklight · Answer

Do you have options for the downward speed?

anonymous · Answer

yes its above the text

anonymous · Answer

find the final velocity and break it into its components (x,y) the use arctangent for angle

anonymous · Answer

intial velocity = 26i

there is no acceleration in horizontl direction.

the initial velocity in vertical direction = 0

after if has fallen let it have a velocity v

v^2 - u^2 = 2*a*s

v^2 = 2 * 10 * 2

v^2 =40

v = 2 * sqrt 10

so the final velocity has its vertical component of 2 * sqrt 10
since there is no acceleration in horizontal direction the horizontal component is 26

so the angle made with horizontal= 

tan (theta) = 2*sqrt 10 / 26

theta = tan^-1 ( sqrt 10 / 13)

anonymous · Answer

is my answer right?