To determine the distance traveled by the cannonball, we can analyze the horizontal and vertical components separately.
The initial velocity in the horizontal direction (Vx) remains constant throughout the motion as there is no force acting in this direction. Therefore, Vx = initial velocity * cos(angle).
Given that the initial velocity is 112 m/s and the angle is 40 degrees, we have:
Vx = 112 * cos(40) ≈ 86 m/s.
The time of flight (t) can be calculated from the vertical component of the motion. The vertical velocity (Vy) changes due to the force of gravity, but it reaches zero at the highest point of the trajectory. Thus, we can determine the time taken to reach the maximum height and then multiply it by 2 to account for the complete round trip of the ball.
The initial vertical velocity (Vy₀) is given by Vy₀ = initial velocity * sin(angle).
Given that the angle is 40 degrees, we have:
Vy₀ = 112 * sin(40) ≈ 71.7 m/s.
The time taken to reach the maximum height can be found using the formula:
Vy = Vy₀ - g*t,
where g is the acceleration due to gravity (approximately 9.8 m/s²).
At the highest point, Vy equals zero, so 0 = Vy₀ - g*t.
Therefore, t = Vy₀ / g.
Using the calculated vertical initial velocity of 71.7 m/s and the acceleration due to gravity of 9.8 m/s², we find:
t = 71.7 / 9.8 ≈ 7.3 s.
Since the time of flight is the same for the upward and downward paths, the total flight time is given by 2t = 2 * 7.3 ≈ 14.6 s.
Finally, to determine the horizontal distance traveled (d), we can multiply the horizontal component of velocity (Vx) by the total flight time (2t):
d = Vx * 2t = 86 * 14.6 ≈ 1255.6 m.
Therefore, the cannonball will travel approximately 1255.6 meters.
the average speed of a cannonball shot out of a standard cannon is 112m/s. if the cannon is titled so the cannonball launches at a 40 degree angle, how far will the cannonball travel?
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