An arrow is fired horizontally from the top of a 40.0-m vertical cliff and lands 125 m away. Therefore the arrow was fired at _____ m/s.

t=sqrt{2h/g}

v=L/t

To find the speed at which the arrow was fired, we can use the concept of projectile motion. When the arrow is fired horizontally, it only experiences motion in the horizontal direction, while the vertical motion is solely due to the force of gravity.

To solve for the speed, we can use the equation:

š‘‘ = š‘£š‘”

Where:
d = distance traveled (125 m)
v = horizontal speed (unknown)
t = time of flight (unknown)

First, we need to find the time of flight. In the vertical direction, we can use the equation:

š‘‘ = 1/2š‘”š‘”Ā²

Where:
d = vertical distance (40 m)
g = acceleration due to gravity (9.8 m/sĀ²)
t = time of flight (unknown)

Rearranging the equation, we have:

š‘” = āˆš(2š‘‘/š‘”)

Substituting the values, we get:

š‘” = āˆš(2 * 40 / 9.8) ā‰ˆ 2.02 s

Now that we have the time of flight, we can substitute it back into the first equation to solve for the horizontal speed:

125 m = š‘£ * 2.02 s

š‘£ = 125 m / 2.02 s ā‰ˆ 61.88 m/s

Therefore, the arrow was fired at approximately 61.88 m/s.