Question

A mug slides off a counter 1.22m high and lands 1.4 meters from the base. With what velocity did the mug leave the counter and what is the direction of the mug's velocity just before it hits the ground?

Answer 1

y = v0t + .5atsquared need to find time it takes to drop y = .5atsquared
1.22 = .5 (9.8) t squared = t= 4.89s plug this into the previous equation sorry about that

Answer 2

i answered the first question

Answer 3

So it travels 1.4m horizontally in the time it takes to fall 1.22m vertically.

s=ut + 1/2 a t^2

1.22 = 0 + 1/2 * 9.81 * t^2 will give you the time

That time and horizontal distance (1.4m) will give you the initial horizontal velocity (which remains the same throughout). You will then need the vertical velocity achieved during the time you got earlier.

v = u +at

The final answer you need is the vector addition of the two velocities. This will be a triangle. You find its length (magnitude) by Pythagoras and then use trig to get the angle.