Final answer:
To calculate the horizontal distance traveled by a cannonball shot at a 60-degree angle with an initial velocity of 25 m/s, we must first find the horizontal velocity component and the time of flight. The horizontal distance is then the horizontal velocity multiplied by the time of flight, which in this case results in approximately 48.75 meters.
Step-by-step explanation:
To solve for the horizontal distance a cannonball travels when fired from ground level at a 60-degree angle with an initial velocity of 25 m/s, we can use the concepts of projectile motion. First, we need to find the horizontal (x) and vertical (y) components of the initial velocity. The horizontal component (Vx) is given by the initial velocity multiplied by the cosine of the angle of projection, and the vertical component (Vy) is given by the initial velocity multiplied by the sine of the angle of projection.
The horizontal component for this problem will be Vx = 25 m/s * cos(60°) = 12.5 m/s. To find the distance traveled, we need the time of flight, which can be found by using the vertical component and solving for the time it takes to go up and come back down to the original launch height (ground level), a complete trajectory. The time of flight is given by T = (2 * Vy) / g, where g is the acceleration due to gravity (9.8 m/s²). So, T = (2 * 25 m/s * sin(60°)) / 9.8 m/s² = about 3.9 seconds.
Finally, to find the horizontal distance traveled (Range), we multiply the horizontal velocity by the time of flight: Range = Vx * T = 12.5 m/s * 3.9 s = 48.75 meters. Thus the cannonball travels approximately 48.75 meters horizontally.