Since the stone is thrown horizontally, its vertical velocity remains constant throughout the motion, assuming no air resistance. The only force acting on the stone in the vertical direction is gravity, which causes it to accelerate downward at a rate of approximately 9.8 m/s² (ignoring air resistance).
In this case, we can calculate the stone's vertical displacement after 4 seconds using the formula:
s = ut + (1/2)at²
Where: s = displacement (vertical) u = initial velocity (vertical) t = time a = acceleration (vertical)
Since the stone is thrown horizontally, its initial vertical velocity (u) is 0 m/s. The acceleration (a) is -9.8 m/s² (negative because it's downward).
Plugging in the values:
s = 0(4) + (1/2)(-9.8)(4²) s = -78.4 meters
So, after 4 seconds, the stone will have fallen 78.4 meters below its initial position.
However, the question asks for the stone's velocity after 4 seconds, not its displacement. Since the vertical velocity remains constant, the stone's velocity after 4 seconds will also be its initial vertical velocity, which is 0 m/s.