A cannonball is shot vertically upward with a velocity of 55m/s. As I delve into this fascinating scenario, I can’t help but wonder about the laws of physics at play here. How high will the cannonball go? What factors will determine its trajectory? Join me as we explore the intricacies of projectile motion and uncover the answers to these questions.

When a cannonball is launched vertically upward, it defies gravity’s pull for a brief moment before succumbing to its inevitable descent. The initial velocity of 55m/s propels the cannonball into the air, defying our expectations and challenging our understanding of motion. But what happens next? Will it reach a towering height or fall back to Earth sooner than anticipated?

By analyzing the forces acting on the cannonball during its ascent, we can begin to unravel its journey through space and time. Factors such as gravitational acceleration, air resistance (negligible in this idealized scenario), and launch angle all come into play. Through mathematical equations and careful calculations, we can estimate how high the cannonball will rise before reversing its course.

Stay tuned as I break down each stage of this intriguing experiment, providing you with insights that bridge theory and practical application. Together, let’s uncover the mysteries behind projectiles in motion and gain a deeper appreciation for the laws that govern their behavior in our universe.

## A Cannonball is Shot Vertically Upward with a Velocity of 55m/s

### Calculating the Maximum Height Reached

When a cannonball is shot vertically upward with a velocity of 55m/s, one of the key aspects to consider is the maximum height it can reach. To calculate this, we can use some basic principles of physics.

Using the equation v^2 = u^2 + 2as, where v represents final velocity, u represents initial velocity, a represents acceleration, and s represents displacement or distance traveled, we can solve for s (in this case, the maximum height). Since the cannonball reaches its highest point when its final velocity becomes zero before falling back down due to gravity’s pull, we can set v to be zero in our equation.

By plugging in the given initial velocity (u = 55m/s) and rearranging the equation to solve for s, we find that:

s = -u^2 / (2a)

Here, “a” denotes acceleration due to gravity which is approximately equal to -9.8 m/s^2. Plugging in these values into our equation gives us:

s = -(55)^2 / (2 * -9.8)

Calculating this expression reveals that the maximum height reached by the cannonball is approximately 151 meters.

### Determining the Time Taken to Reach the Highest Point

Another factor worth exploring is how long it takes for the cannonball to reach its highest point after being shot vertically upward. To determine this time interval, we need to utilize kinematic equations.

The formula that relates initial velocity (u), final velocity (v), acceleration (a), and time taken (t) is:

v = u + at

Since we want to find out when v becomes zero at its highest point of trajectory, we can rewrite this equation as:

0 = u + at

Rearranging and solving for t yields:

t = -u / a

Substituting the given values of u = 55m/s and a = -9.8m/s^2, we can calculate the time taken to reach the highest point:

t = -(55) / (-9.8)

Hence, it takes approximately 5.61 seconds for the cannonball to reach its maximum height.

## Analyzing the Cannonball’s Velocity at Different Points

To gain further insight into the motion of the cannonball, let’s examine its velocity at various points along its trajectory. As the ball moves upward against gravity’s pull, its velocity decreases until it reaches zero at the highest point. After that, it starts descending back down with an increasing velocity due to acceleration from gravity.

At any point during this journey, we can determine its velocity using another kinematic equation:

v = u + at

Where v represents final velocity, u denotes initial velocity, a stands for acceleration (in this case -9.8 m/s^2), and t signifies time elapsed.

For instance, if we want to find out the velocity of the cannonball after 3 seconds from launch when moving upward against gravity’s force, we can plug in u = 55m/s and t = 3s into our equation:

v = 55 – (9.8 * 3)

Calculating this expression reveals that after 3 seconds of ascent, the cannonball has a velocity of approximately 16.6 m/s.

By applying similar calculations at different time intervals or points along its trajectory, we can obtain valuable information about how the cannonball’s speed changes as it ascends and descends in vertical motion.