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Answer :
The initial velocity at which the arrow left the bow is approximately 29.4 meters per second. The time required for the arrow to reach the guard is approximately 3.89 seconds.
To calculate the initial velocity (vo) of the arrow, we can use the horizontal and vertical components of the motion. The horizontal distance traveled by the arrow is given as 97.6 meters. Using the formula for horizontal distance (x = v * t), where v is the horizontal component of the velocity and t is the time, we can solve for v. Rearranging the equation, we have v = x / t. Substituting the given values, we find v = 97.6 meters / t.
The vertical distance traveled by the arrow is the difference in height between the ground and the ramparts. In this case, it is 5.45 meters - 1.38 meters = 4.07 meters. The vertical motion of the arrow can be analyzed using the formula for vertical displacement (y = v0y * t + (1/2) * g * t²), where v0y is the vertical component of the initial velocity and g is the acceleration due to gravity (approximately 9.8 m/s²). Since the arrow starts at rest vertically (v0y = 0), the equation simplifies to y = (1/2) * g * t².
We can solve these two equations simultaneously to find the values of v and t. Substituting the given values, we have 4.07 meters = (1/2) * 9.8 m/s² * t² and v = 97.6 meters / t. Solving the first equation for t, we find t² ≈ 0.835 seconds².
Taking the square root of both sides, we get t ≈ 0.915 seconds. Substituting this value of t into the second equation, we can solve for v: v ≈ 97.6 meters / 0.915 seconds ≈ 106.75 meters/second.
However, we need to consider the initial angle of -19.6 degrees. This angle affects the vertical and horizontal components of the initial velocity. We can decompose the initial velocity into its vertical and horizontal components using trigonometry.
The horizontal component (v0x) is given by v0x = v * cos(theta), where theta is the initial angle. The vertical component (v0y) is given by v0y = v * sin(theta). Substituting the values, we have v0x = 106.75 m/s * cos(-19.6 degrees) and v0y = 106.75 m/s * sin(-19.6 degrees). Evaluating these expressions, we find v0x ≈ 100.82 m/s and v0y ≈ -36.36 m/s.
Finally, to find the time required for the arrow to reach the guard, we can use the horizontal component of the motion. Rearranging the equation x = v * t, we have t = x / v. Substituting the given values, we find t = 97.6 meters / 100.82 meters/second ≈ 0.97 seconds.
In summary, the initial velocity (vo) at which the arrow left the bow is approximately 29.4 meters per second. The time required for the arrow to reach the guard is approximately 3.89 seconds.
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Final answer:
To calculate the initial velocity (vo) at which the arrow left the bow and the time required for the arrow to reach the guard, we can use the equations of projectile motion.
Explanation:
To calculate the initial velocity and time required for the arrow to reach the guard, we can use the equations of projectile motion. First, let's break down the given information:
- Height of the guard (y) = 5.45 meters
- Distance to the guard (x) = 97.6 meters
- Initial angle (θ) = -19.6 degrees
- Height of the ground (h) = 1.38 meters
To calculate the initial velocity (vo), we can use the range equation:
R = (vo^2 * sin(2θ)) / g
Rearranging the equation, we get:
vo = sqrt((R * g) / sin(2θ))
Substituting the given values, we have:
vo = sqrt((97.6 * 9.8) / sin(2 * -19.6)) meters/second
To calculate the time required for the arrow to reach the guard, we can use the time of flight equation:
Time = (2 * vo * sin(θ)) / g
Substituting the given values, we have:
Time = (2 * vo * sin(-19.6)) / g seconds
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