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

`8.72s`

Step-by-step explanation

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

`T=(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=(2\cdot85.5\cdot\sin 30)/(9.8) \\ T=8.72s \end{gathered}`