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A cannonball is launched upward with a velocity of 85.5m/s at an angle of 30 degrees above the horizontal.How long is the cannonball in the air?

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ANSWER

[tex]8.72s[/tex]

EXPLANATION

To find the time of flight, apply the formula for a projectile's time of flight:

[tex]T=\frac{2u\sin \theta}{g}[/tex]

where u = initial velocity

θ = angle of launch

g = acceleration due to gravity

Therefore, the time of flight of the cannonball is:

[tex]\begin{gathered} T=\frac{2\cdot85.5\cdot\sin 30}{9.8} \\ T=8.72s \end{gathered}[/tex]

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