Question

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?

Answer 1

ANSWER

$8.72s$

EXPLANATION

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

$T=\frac{2u\sin \theta}{g}$

where u = initial velocity

θ = angle of launch

g = acceleration due to gravity

Therefore, the time of flight of the cannonball is:

$\begin{gathered} T=\frac{2\cdot85.5\cdot\sin 30}{9.8} \\ T=8.72s \end{gathered}$