Question

A hockey puck slides off the edge of a horizontal platform with an initial velocity of 20 m/s horizontally and experiences no significant air resistance. The height of the platform above the ground is 2.0 m. What is the magnitude of the vertical component of the velocity of the puck just before it hits the ground

Answer 1

`v_(yf)=6.26m/s`

Step-by-step explanation:

We have to focus only on the vertical direction: the puck falls 2m with a null vertical initial velocity. We use the equation:

`v_f^2=v_i^2+2ad`

which for the vertical direction is:

`v_(yf)^2=v_(yi)^2+2a_y\Delta y`

or:

`v_(yf)=\sqrt{v_(yi)^2+2a_y\Delta y}`

Taking the downward direction as positive (which means a positive gravitational acceleration and a positive displaceent) we have for our values:

`v_(yf)=√((0m/s)^2+2(9.8m/s^2)(2m))=6.26m/s`