A cannon fires a cannonball 500 m downrange when sest at a 45° angle. At what velocity does the cannonball leave the cannon?
To find the velocity at which the cannonball leaves the cannon, we can use the horizontal and vertical components of the initial velocity.
Let's assume:
- The initial velocity of the cannonball is V.
- The vertical component of the initial velocity is Vy.
- The horizontal component of the initial velocity is Vx.
Since the cannonball is fired at a 45° angle, we know that Vy = Vx.
We can use the equations of motion to find the components of the initial velocity.
For the horizontal component:
Vx = V * cos(45°)
For the vertical component:
Vy = V * sin(45°)
We can substitute Vy = Vx into the equations above:
Vy = Vx = V * sin(45°)
Now, we need to find the time it takes for the cannonball to reach the maximum height (when it vertically reaches its highest point). At this point, the vertical velocity becomes zero.
Using the formula for displacement in vertical motion:
Vy = Vy0 + a * t
Since the vertical velocity at the maximum height is zero (Vy = 0), we can rewrite the equation as:
0 = Vy0 + (-9.8 m/s^2) * t
Solving for t:
t = Vy0 / 9.8
Now, we can use the time to find the total time of flight (time taken to reach the maximum height and then return back to the ground):
Total time = 2 * t
Using the time and horizontal component of the velocity, we can find the distance traveled horizontally:
Distance = Vx * Total time
Given that the distance traveled is 500 m horizontally, we can rewrite the equation above as:
500 m = V * cos(45°) * 2 * (Vy0 / 9.8)
Since Vy0 = V * sin(45°), we can substitute it into the equation above:
500 m = V * cos(45°) * 2 * ((V * sin(45°)) / 9.8)
Simplifying the equation:
500 m = V^2 * 2 * sin(45°) * cos(45°) / 9.8
Or:
500 m = V^2 * sin(2 * 45°) / 9.8
Solving for V^2:
V^2 = (500 m * 9.8) / sin(2 * 45°)
Taking the square root of both sides:
V = √((500 m * 9.8) / sin(2 * 45°))
Calculating the result:
V ≈ √(4900 m^2/s^2 / 1)
V ≈ √(4900 m^2/s^2)
V ≈ 70 m/s
Therefore, the velocity at which the cannonball leaves the cannon is approximately 70 m/s.
To find the velocity at which the cannonball leaves the cannon, we can use basic principles of projectile motion.
We are given that the cannonball travels a distance of 500 meters downrange, and it is launched at an angle of 45 degrees. Here, we assume that there is no air resistance.
The motion of the projectile can be divided into horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component is influenced by gravity.
1. Horizontal Component:
The horizontal distance traveled (range) can be calculated using the horizontal component of velocity (Vx) and the time of flight (t). Since there is no acceleration horizontally, the horizontal velocity remains constant.
2. Vertical Component:
The vertical distance traveled can be calculated using the initial vertical velocity (Vy) and time of flight (t). The vertical velocity changes due to the effect of gravity.
To find the initial velocity (V) at which the cannonball leaves the cannon, we need to find the horizontal and vertical components of velocity using the given information.
1. Horizontal Component:
The horizontal component of velocity (Vx) can be found using the formula Vx = V * cosθ, where θ is the angle of projection. In this case, the angle of projection is 45 degrees, so cosθ = cos45 = 0.7071 (rounded to four decimal places).
2. Vertical Component:
The initial vertical velocity (Vy) can be found using the formula Vy = V * sinθ. Again, since the angle of projection is 45 degrees, sinθ = sin45 = 0.7071.
Now, we have the horizontal component (Vx) and vertical component (Vy) of the initial velocity.
3. Time of Flight:
The time of flight (t) can be calculated using the vertical component of motion. The formula to find the time of flight is t = 2 * Vy / g, where g is the acceleration due to gravity (approximately 9.8 m/s²).
Using the given vertical component value (Vy = V * sinθ) and the equation t = 2 * Vy / g, we can calculate the time of flight (t).
4. Range:
The range (horizontal distance) can be calculated using the horizontal component of velocity (Vx) and the time of flight (t). The formula for range is R = Vx * t.
Using the given horizontal component value (Vx = V * cosθ) and the calculated time of flight value (t), we can calculate the range (R).
Now, using the given range value (R = 500 m) and the calculated time of flight value (t), we can solve for the initial velocity (V).
Once you have the values for velocity (Vx and Vy), time of flight (t), and range (R), you can substitute them into the equations mentioned earlier to calculate the initial velocity (V) at which the cannonball leaves the cannon.