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Answer :
The velocity of the fireworks shell when it leaves the mortar can be calculated using the principles of projectile motion. In this case, we can neglect air resistance, although it is important to note that in reality, air resistance would have an impact on the shell's velocity.
To find the shell's velocity when it leaves the mortar, we can use the equation:
[tex]v^2 = u^2 + 2as[/tex]
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Since the shell is fired straight up, the initial velocity is zero (u = 0). The acceleration due to gravity, a, is approximately [tex]9.8 m/s^2[/tex] (assuming we are near the surface of the Earth).
The displacement, s, is the height the shell reaches, which is given as 197 m.
Plugging these values into the equation, we have:
[tex]v^2 = 0^2 + 2 * 9.8 * 197[/tex]
Simplifying the equation, we get:
[tex]v^2 = 2 * 9.8 * 197[/tex]
[tex]v^2 = 3848.4[/tex]
Taking the square root of both sides, we find:
v ≈ 62.06 m/s
Therefore, the magnitude of the shell's velocity when it leaves the mortar is approximately 62.06 m/s.
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