119) You throw a rock horizontally off a cliff with a speed of 20 m/s and no significant air resistance. After 2.0 s, the magnitude of the velocity of the rock is closest to

44.54 m/s

Explanation:

Let's break the problem into horizontal and vertical.

Horizontal:

Here we do not have any force, only a constant horizontal velocity, so we can write:

Vx(t) = 20m/s.

Vertical:

Here we have the gravitational acceleration acting on the rock, then we can write:

Ay(t) = -9.8m/s^2

For the vertical velocity, we need to integrate over time and get:

Vy(t) = (-9.8m/s^2)*t + V0

Where V0 is the initial vertical velocity, but the rock was thrown horizontally, so we do not have an initial velocity in the vertical axis.

Vy(t) = (-9.8m/s^2)*t.

Ok, now we want to know the magnitude of the velocity when t = 2.0s.

The magnitude will be equal to:

V = √( Vx(2s)^2 + Vy(2s)^2)

   = √(  (20m/s)^2 + (-2s*9.8m/s^2)^2) = 44.54 m/s

The magnitude of the velocity after 2 seconds is around 44.54 m/s

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5

he kicked it 10 feet straight up into the air? i think this works.