A cannon sits on a wall such that the barrel of the cannon is 12 meters above the ground. The target is 220 meters away. In order to hit the target, what initial velocity must the cannonball have in m/s?
To solve this problem, we can use the principles of projectile motion. The cannonball launched from the cannon will follow a curved trajectory known as a projectile.
The key here is to recognize that the horizontal and vertical motions of the cannonball are independent of each other. This means we can analyze the vertical and horizontal components separately.
To find the initial velocity required, we need to determine the horizontal and vertical components of the velocity.
Step 1: Determine the time of flight
The horizontal distance traveled by the cannonball (220 meters) is equal to the horizontal velocity multiplied by the time of flight. However, since there is no acceleration in the horizontal direction, the horizontal velocity remains constant throughout the motion.
Step 2: Determine the vertical displacement
The vertical displacement of the cannonball is the difference between the initial height (12 meters) and the final height (ground level, 0 meters).
Step 3: Use the kinematic equations
Using the vertical motion equations, we can find the time of flight and the required initial vertical velocity component.
The vertical displacement is given as -12 meters (negative because we are measuring downwards), the initial vertical velocity component is unknown (let's call it Vy), the acceleration due to gravity is -9.8 m/s² (negative because it acts downward), and the time of flight is also unknown (let's call it T).
Vertical displacement equation:
h = Viy × T + (1/2) × (-9.8) × T²
Since the cannonball lands on the ground, the final height is 0, so we can solve for T.
0 = Viy × T + (1/2) × (-9.8) × T²
Step 4: Solve for T
Using the quadratic formula or factoring, we can solve for T. In this case, let's use the quadratic formula.
0 = (1/2) × (-9.8) × T² + Viy × T
Simplifying the equation and setting it equal to zero:
4.9T² + Viy × T = 0
Using the quadratic formula:
T = (-b ± √(b² - 4ac)) / (2a)
Here: a = 4.9, b = Viy, and c = 0
T = (-Viy ± √(Viy² - 4 × 4.9 × 0)) / (2 × 4.9)
T = (-Viy ± √(Viy²)) / 9.8
T = (-Viy ± Viy) / 9.8
Since the time of flight cannot be negative, we discard the negative value:
T = (-Viy + Viy) / 9.8
T = (0) / 9.8
T = 0
Step 5: Velocity in the horizontal direction
Using the formula v = d / t, where v is the horizontal velocity and d is the horizontal displacement, we can find the horizontal velocity.
v = 220 / T
Since T = 0, the horizontal velocity is undefined.
Step 6: Final calculation of initial velocity
To find the initial velocity of the cannonball, we need to use the Pythagorean theorem because the initial velocity is the vector sum of the horizontal and vertical initial velocities.
v_initial = √(v_horizontal² + v_vertical²)
However, since the horizontal velocity is undefined in this case, the initial velocity will only have a vertical component.
v_initial = |v_vertical|
Therefore, the initial velocity required for the cannonball to hit the target is equal to the magnitude of the vertical velocity component.
Conclusion:
To find the initial velocity, we need to find the vertical velocity component, which we calculated as 0. Therefore, the cannonball would need to be shot vertically (straight up or straight down) in order to hit the target.