Answer:
49.2 m/s
Explanation:
[tex]\boxed{\begin{minipage}{9 cm}\underline{The Constant Acceleration Equations (SUVAT)}\\\\s = displacement in m (meters)\\u = initial velocity in m s$^{-1}$ (meters per second)\\v = final velocity in m s$^{-1}$ (meters per second)\\a = acceleration in m s$^{-2}$ (meters per second per second)\\t = time in s (seconds)\\\\When using SUVAT, assume the object is modeled\\ as a particle and that acceleration is constant.\end{minipage}}[/tex]
Consider the vertical and horizontal motion of the cannonball separately.
Vertical component of velocity
If the cannonball is shot horizontally, the vertical component of its initial velocity (u) is zero.
Because the projectile is modeled as moving only under the influence of gravity, the only acceleration the projectile will experience will be acceleration due to gravity (a = 9.8 m/s²).
To calculate the vertical component of the cannonball's velocity at 1.23 s, resolve vertically, taking down as positive:
[tex]u = 0, \quad v = v_y, \quad a = 9.8, \quad t = 1.23[/tex]
[tex]\begin{aligned}\textsf{Using }\:\: v & = u+at\\\\\implies v_y & =0+9.8 \times 1.23\\& = 12.054 \:\: \sf m/s\end{aligned}[/tex]
Horizontal component of velocity
The horizontal component of velocity is constant, as there is no acceleration horizontally, so:
[tex]v_x = u_x = 47.7 \:\: \sf m/s[/tex]
Magnitude of velocity
To find the magnitude of the cannonball's velocity:
[tex]\begin{aligned} |v| & = \sqrt{{v_x}^2+{v_y}^2}\\& = \sqrt{47.7^2+(-12.054)^2}\\& = 49.19948085...\\& = 49.2 \:\: \sf m/s\:\: (3\: s.f.)\end{aligned}[/tex]
Therefore, the magnitude of the cannonball's velocity after 1.23 s is 49.2 m/s.
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