Answer: 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
