Question

When launching a model cannon, students measure the height of the launch point to be 15 m and the horizontal distance the cannonball traveled to be 82 m. Assuming the cannon was fired horizontally what was the launch velocity of the cannonball ?

Answer 1

The launch velocity of the cannonball fired horizontally from a height of 15 m to cover a horizontal distance of 82 m is approximately 46.83 m/s, determined using the equations of projectile motion for both vertical and horizontal displacement.

Step-by-step explanation:

To determine the launch velocity of the cannonball fired horizontally from a height of 15 m and travelling a horizontal distance of 82 m, we can use the physics of projectile motion. Since the cannonball was fired horizontally, its initial vertical velocity is 0 m/s, and the only acceleration it experiences is due to gravity (9.81 m/s2). The time it takes to fall from the launch height can be derived using the formula for vertical displacement y = (1/2) * g * t2, where y is the vertical distance, g is the acceleration due to gravity, and t is the time in seconds.

First, solve for time (t) using the vertical motion equation y = (1/2) * g * t2. Plugging in the height of 15 m:

15 m = (1/2) * 9.81 m/s2 * t2

This gives us a time of approximately 1.7507 seconds. Then, the launch velocity (vx) is calculated using the horizontal distance x = vx * t, where x is 82 m. Solving for vx:

82 m = vx * 1.7507 s

The launch velocity vx is approximately 46.83 m/s.