A cannonball is shot with initial speed 230m/s at a 50-degree angle above the horizontal. Assuming air resistance is negligible, what is the acceleration of the cannonball when it is at its maximum height?
To determine the acceleration of the cannonball at its maximum height, we need to consider the motion of the projectile in two separate directions: horizontal and vertical.
First, let's calculate the initial velocity components of the cannonball. The initial speed of 230 m/s can be broken down into horizontal and vertical components using trigonometric relations.
The vertical component is given by V₀y = V₀ * sin(θ), where V₀ is the initial speed and θ is the angle above the horizontal. So, V₀y = 230 m/s * sin(50°).
The horizontal component is given by V₀x = V₀ * cos(θ), where V₀ is the initial speed and θ is the angle above the horizontal. So, V₀x = 230 m/s * cos(50°).
At the maximum height, the vertical component of velocity becomes zero since the projectile reaches its highest point and starts falling back down.
Now, let's consider the vertical motion. The only force acting on the cannonball in the vertical direction is gravity, which causes acceleration downward.
The acceleration due to gravity on Earth is approximately 9.8 m/s², always pointing downward. Therefore, at the maximum height, the acceleration of the cannonball is equal to the acceleration due to gravity, which is 9.8 m/s².
So, the acceleration of the cannonball at its maximum height is 9.8 m/s².
To find the acceleration of the cannonball when it is at its maximum height, we need to consider that at the highest point of the trajectory, the vertical component of the velocity will be zero, while the horizontal component of velocity remains constant.
1. Split the initial velocity into its vertical and horizontal components:
Vertical component: v_y = v_initial * sin(theta)
Horizontal component: v_x = v_initial * cos(theta)
where:
v_initial = initial velocity of the cannonball (230 m/s)
theta = angle above the horizontal (50 degrees)
2. At the maximum height, the vertical velocity becomes zero, so we can find the time it takes for the cannonball to reach this point using the vertical motion equation:
v_y = v_initial * sin(theta) - g * t_max_height = 0
Simplifying the equation, we get:
t_max_height = (v_initial * sin(theta)) / g
where:
g = acceleration due to gravity (approximately 9.8 m/s^2)
3. Finally, we can calculate the acceleration at the maximum height by finding the horizontal acceleration, which remains constant throughout the projectile motion:
a = -g
Substituting the values:
a = -9.8 m/s^2
Therefore, the acceleration of the cannonball when it is at its maximum height is approximately -9.8 m/s^2 (downward).