A cannonball is fired with a velocity of 200 m/s at an angle of 25 degrees above the ground. After the total flight of 45.6 seconds, the cannonball returns to the ground. Calculate: the initial horizontal and vertical components of the velocity, the maximum height of the cannonball above the ground, and the total horizontal range.
reyrer
500
To solve this problem, we can break the initial velocity into its horizontal and vertical components.
Step 1: Calculate the initial horizontal component of the velocity.
The horizontal component of the velocity (Vx) remains constant throughout the motion. It is given by the formula:
Vx = V * cos(theta)
where:
V = initial velocity = 200 m/s
theta = angle above the ground = 25 degrees
Substituting the values, we have:
Vx = 200 m/s * cos(25 degrees)
Using a scientific calculator, calculate cos(25 degrees) and multiply it by 200 m/s to find Vx.
Step 2: Calculate the initial vertical component of the velocity.
The vertical component of the velocity (Vy) changes throughout the motion. At the highest point of the trajectory, Vy is zero. We can calculate the initial vertical component of the velocity using the formula:
Vy = V * sin(theta)
where:
V = initial velocity = 200 m/s
theta = angle above the ground = 25 degrees
Substituting the values, we have:
Vy = 200 m/s * sin(25 degrees)
Using a scientific calculator, calculate sin(25 degrees) and multiply it by 200 m/s to find Vy.
Step 3: Calculate the time of flight.
The time of flight for the projectile is given by the formula:
Time of flight = 2 * Vy / g
where:
Vy = initial vertical component of the velocity
g = acceleration due to gravity = 9.8 m/s^2
Substituting the values, we have:
Time of flight = 2 * Vy / g
Step 4: Calculate the maximum height.
The maximum height reached by the projectile can be found using the formula:
Maximum height = Vy^2 / (2 * g)
where:
Vy = initial vertical component of the velocity
g = acceleration due to gravity = 9.8 m/s^2
Substituting the values, we have:
Maximum height = Vy^2 / (2 * g)
Step 5: Calculate the total horizontal range.
The horizontal range is the distance covered horizontally by the projectile during its flight. It can be calculated using the formula:
Horizontal range = Vx * Time of flight
where:
Vx = initial horizontal component of the velocity
Time of flight = total flight time
Substituting the values, we have:
Horizontal range = Vx * Time of flight
Now, substitute the calculated values into the respective formulas to get the final answers.
To solve this problem, we can break the initial velocity into its horizontal and vertical components.
1. Initial Horizontal Component of Velocity:
The horizontal component of the velocity remains constant throughout the flight because there is no horizontal acceleration. Given the initial velocity of 200 m/s and the launch angle of 25 degrees, we can calculate the horizontal component of velocity using trigonometry.
Horizontal Velocity (Vx) = Initial Velocity (V) * cos(angle)
Vx = 200 m/s * cos(25 degrees)
Vx ≈ 200 m/s * 0.9063
Vx ≈ 181.26 m/s
Therefore, the initial horizontal component of velocity is approximately 181.26 m/s.
2. Initial Vertical Component of Velocity:
The vertical component of the velocity changes due to gravity. We need to calculate the vertical component of velocity at the highest point of the trajectory. At the highest point, the vertical velocity component becomes zero. We can calculate the vertical component of velocity using the time of flight and acceleration due to gravity.
Time of Flight (T) = 45.6 seconds
Vertical Velocity (Vy) = -g * T/2
(gravity acts in the downward direction, so we use a negative sign)
Vy = -9.8 m/s^2 * 45.6 s/2
Vy ≈ -222.48 m/s
Therefore, the initial vertical component of velocity is approximately -222.48 m/s.
3. Maximum Height:
To find the maximum height, we need to find the time at which the cannonball reaches the highest point. At the highest point, the vertical velocity becomes zero. We can calculate the time of flight to reach the highest point using the initial vertical component of velocity and acceleration due to gravity.
0 = Vy - g * t
t = Vy / g
t = -222.48 m/s / 9.8 m/s^2
t ≈ -22.71 seconds
Since time cannot be negative, we take the absolute value.
t ≈ 22.71 seconds
Now, we can calculate the maximum height using the time at the highest point and the vertical component of velocity.
Maximum Height (H) = Vy * t + (1/2) * g * t^2
H = -222.48 m/s * 22.71 s + (1/2) * 9.8 m/s^2 * (22.71 s)^2
H ≈ -5054.82 m
Since height cannot be negative, we take the absolute value.
H ≈ 5054.82 m
Therefore, the maximum height of the cannonball above the ground is approximately 5054.82 meters.
4. Total Horizontal Range:
The horizontal range is the total distance traveled by the cannonball horizontally before it hits the ground. We can calculate the horizontal range using the initial horizontal component of velocity and the total time of flight.
Horizontal Range (R) = Vx * T
R = 181.26 m/s * 45.6 s
R ≈ 8268.10 m
Therefore, the total horizontal range of the cannonball is approximately 8268.10 meters.