Ask AI: A cannon shoots a 200Kg cannonball straight up into the air. The cannonball leaves the cannon at 25m/s. How high will the cannonball travel?

To solve this problem, we can use the basic principles of kinematics. We need to determine how high the cannonball will travel before gravity decelerates it to a stop. This is a projectile motion problem, where we'll use conservation of energy or kinematic equations. Let's use the kinematic equation that relates the final velocity, initial velocity, acceleration, and displacement: \[ v^2 = u^2 + 2as \] Where: - \( v \) is the final velocity (0 m/s, since the cannonball will stop at the peak of its flight), - \( u \) is the initial velocity (25 m/s, the speed at which the cannonball leaves the cannon), - \( a \) is the acceleration due to gravity (-9.81 m/s², negative because it is acting opposite to the direction of motion), - \( s \) is the displacement, or the maximum height reached by the cannonball. Plugging in the known values: \[ 0 = (25)^2 + 2(-9.81)s \] Solving for \( s \): \[ 0 = 625 - 19.62s \] \[ 19.62s = 625 \] \[ s = \frac{625}{19.62} \approx 31.86 \text{ meters} \] Thus, the cannonball will reach a maximum height of approximately 31.86 meters.

The height of the cannonball can be calculated using the formula for the maximum height in a projectile motion which is given by h = (v^2) / (2 * g), where: - v is the initial velocity of the projectile - g is the acceleration due to gravity (on Earth, usually around 9.8 m/s^2). So, if you plug in the given values into the formula: h = (25 m/s)^2 / (2 * 9.8 m/s^2) h = 625 m^2/s^2 / 19.6 m/s^2 h = 31.88 meters So, the cannonball will reach a maximum height of approximately 31.88 meters. Remember that this is an idealized calculation; in real life, air resistance and other factors could cause the actual height to be somewhat less.