A canon placed at the top of a 15-foot rise fires a 16 pound ball with a vertical velocity of 172 feet per second. How far from the canon will the ball land? What is the max height of its trajectory?
This depends on the angle from the horizontal the cannon is fired.
how do i figure that out by the info given?
To find the distance from the cannon where the ball will land, we can use the horizontal velocity of the ball and the time it takes to reach the ground. Since there is no horizontal acceleration, the horizontal velocity remains constant throughout the ball's trajectory.
To determine the horizontal velocity, we need to split the initial velocity into its horizontal and vertical components. The horizontal component of velocity remains unchanged throughout the trajectory. The vertical component of velocity can be determined using the initial vertical velocity.
Given:
Initial vertical velocity (Vy) = 172 ft/s
The time it takes for the ball to reach the ground can be found using the formula:
Time (t) = (2 * Vy) / g
where g is the acceleration due to gravity (32 ft/s^2).
In this case, the time (t) would be:
t = (2 * 172) / 32 = 10.75 seconds (approx.)
To find the horizontal distance traveled (x), we can multiply the time (t) by the horizontal velocity (Vx).
To find the horizontal velocity (Vx), we can use the formula:
Vx = initial horizontal velocity (also written as V0x) = constant
Now, to find the initial horizontal velocity (V0x), we can use the formula:
V0x = V * cos(theta)
where V is the initial velocity of the ball (172 ft/s), and theta is the angle of projection. Since the angle of projection is not given in the question, we assume that the ball is fired horizontally, so the angle is 0 degrees.
Hence, V0x = 172 * cos(0) = 172 ft/s
Now, we can calculate the horizontal distance traveled (x) using the formula:
x = V0x * t
Substituting the values:
x = 172 * 10.75 = 1849 ft (approx.)
Therefore, the ball will land approximately 1849 feet away from the cannon.
To find the maximum height of the trajectory, we can use the formula for vertical motion:
h = (Vy^2) / (2 * g)
Substituting the values:
h = (172^2) / (2 * 32) = 1472 ft (approx.)
Therefore, the maximum height of the trajectory is approximately 1472 feet.