So, it's the angle of impact, not the angle below the table after initial release? Isn't that the angle above the horizontal though? Sorry to be bothersome, but do you mind explaining how to know which angle of reference is to be found by the wording when doing these types of equations?
This is the displacement equation in the horizontal: d=Vi(t)+(1/2)at^2
Since a=0 in the horizontal in cancels (1/2)at^2. I am trying to solve for θ, but this is seemingly the only equation that fits. I feel as though I'm overcomplicating this.
I'm stuck again, frustrating to be stuck on a simple question like this. Time calculated to be 0.63s
d=20cos(theta)(0.63)
I'm missing 2 values, how do i solve it from here on?
I meant I did not have the values required to do the quadratic, no need for the ignorant comment. There's what I wanted to know, initial vertical velocity is 0m/s. Thank you, I get confused with free fall off a platform and projectiles.
I thought there is a horizontal velocity? I'm unsure what you mean, if I use the horizontal value then I do not have the displacement for it. I'm attempting to find the time using the vertical initial velocity, but i have two missing variables: theta and t. I can't do a quadratic. I'm unsure if...
Homework Statement
A hockey puck slides off the edge of a table with an initial velocity of 20 m/s. The height of the table above the ground is 2 m. At what angle below the horizontal does the puck hit the ground?
Homework Equations
d=Vi(t)+(1/2)at^2
a^2+b^2=c^2
The Attempt at a Solution...