To solve this problem, we can use the principles of projectile motion.
Given: Initial vertical velocity (Vy) = 0 m/s (since the arrow is fired horizontally) Vertical displacement (Δy) = -29.4 m (negative because the arrow is moving downward) Horizontal velocity (Vx) = 49.2 m/s
We can calculate the time it takes for the arrow to hit the ground using the vertical motion equation:
Δy = Vy * t + (1/2) * g * t^2
Since Vy is 0, the equation simplifies to:
Δy = (1/2) * g * t^2
Solving for time (t):
t^2 = (2 * Δy) / g
t = sqrt((2 * Δy) / g)
Where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
Substituting the given values:
t = sqrt((2 * -29.4) / 9.8) t = sqrt(-5.96) (Note: The negative sign indicates that we are taking the downward direction as negative.)
Since we're interested in the horizontal velocity, we can use the horizontal motion equation:
Vx = Δx / t
Where Δx is the horizontal displacement, which we don't know yet.
Rearranging the equation to solve for Δx:
Δx = Vx * t
Substituting the given values:
Δx = 49.2 * t
Now, we can substitute the value of t that we calculated earlier:
Δx = 49.2 * sqrt(-5.96)
However, the result is imaginary because the square root of a negative number cannot be directly determined. This implies that the arrow does not hit the ground under the given conditions. Please double-check the values or provide additional information if needed.