An archer shoots and arrow horizontally at a target 15m away. The arrow is aimed directly at the center of the target, but it hits 52 cm lower. what will be its initial velocity.
To find the initial velocity of the arrow, we can use the concept of projectile motion.
In projectile motion, an object is thrown at an angle or launched horizontally, experiencing motion in both the horizontal and vertical directions simultaneously. In this case, the arrow is shot horizontally, which means it only has a horizontal velocity (Vx) and no vertical velocity (Vy) initially. The only force acting on the arrow is gravity, which pulls it downward.
Given that the arrow hits the target 52 cm (or 0.52 m) lower, we have the vertical displacement (Δy) of the arrow. In this case, Δy = -0.52 m because it is below the initial position.
We know that the horizontal displacement (Δx) is 15 m since the arrow hits the target 15 m away. Since the arrow is shot horizontally, the horizontal velocity (Vx) remains constant throughout its motion.
We also know that the acceleration due to gravity (g) is approximately 9.8 m/s², acting downward.
Now, let's find the initial velocity (Vo), which is the magnitude of the velocity vector of the arrow at the moment it is shot.
Since there is no initial vertical velocity (Vy), we can use the vertical motion equation:
Δy = Vy * t + (1/2) * g * t²
Since the arrow is aimed directly at the center of the target, the time taken to reach the target is the same for both the horizontal and vertical components of motion. Therefore, we can multiply the time (t) by the horizontal velocity (Vx) to find the horizontal displacement (Δx). This gives us the relationship:
Δx = Vx * t
Since Vx is constant and t is the same in both equations, we can set Δy = -0.52 m equal to Δx = 15 m to solve for time (t).
Vx * t = Δx
Vx * t = 15 m
Next, we solve the vertical motion equation for time (t):
Δy = Vy * t + (1/2) * g * t²
-0.52 m = 0 * t + (1/2) * (9.8 m/s²) * t²
-0.52 m = 4.9 m/s² * t²
Solving for t:
t² = -0.52 m / (4.9 m/s²)
t² ≈ -0.106122
(Note: We get a negative result because the arrow is below the initial position. We can disregard the negative sign since we are working with magnitudes.)
Now, we can substitute the value of t² back into our previous equation to find Vx:
Vx * t ≈ 15 m
Vx * √(-0.106122) ≈ 15 m
Vx ≈ 15 m / √(-0.106122)
Since square roots of negative numbers are not valid in the real number system, it appears that an error has occurred in the given values or calculations. Please recheck the question and values provided.