Answer:
49.2 m/s
Explanation:
Consider the vertical and horizontal motion of the cannonball <u>separately</u>.
<h3><u>
Vertical component of velocity</u></h3>
If the cannonball is shot horizontally, the <u>vertical component</u> of its initial velocity (u) is zero.
Because the projectile is modeled as <u>moving</u> only under the influence of gravity, the only acceleration the projectile will experience will be <u>acceleration due to gravity</u> (a = 9.8 m/s²).
To calculate the <u>vertical component</u> of the cannonball's velocity at 1.23 s, <u>resolve vertically</u>, taking <u>down as positive</u>:
<h3><u>Horizontal component of velocity</u></h3>
The <u>horizontal component</u> of velocity is constant, as there is no acceleration horizontally, so:
<h3><u>Magnitude of velocity</u></h3>
To find the magnitude of the cannonball's velocity:
Therefore, the magnitude of the cannonball's velocity after 1.23 s is 49.2 m/s.
Learn more about constant acceleration equations here:
brainly.com/question/26241670
brainly.com/question/27976125