A cannon ball is fired horizontally from a cannon that is 40.7m above the ground. The initial velocity of the cannon ball is 234m/s. How much time does the cannon ball spend in the air?

how much time does it take to fall?

h=1/2 g t^2
t=sqrt(2*40.7/9.8) do that.
Now, how far does it go horizontally?
distance=234m/s * timeinAir

t= sqrt (8.306) = 2.88s

d= 234m/s * 2.88s = 674.40m

Thanks!

To find the time it takes for the cannonball to be in the air, we can use the kinematic equation:

h = v₀t + (1/2)gt²

Where:
h = height (in this case, 40.7m)
v₀ = initial velocity (234m/s)
t = time
g = acceleration due to gravity (-9.8m/s², assuming a downward direction)

Since the cannonball is fired horizontally, its initial vertical velocity is 0m/s.

So, the equation becomes:

40.7 = 0 * t + (1/2) * (-9.8) * t²

Simplifying the equation:

40.7 = -4.9t²

Rearranging the equation to solve for t:

t² = 40.7 / -4.9

t² ≈ -8.327

Since time cannot be negative, we discard the negative value. Hence, the equation has no real solution.

This result indicates that the cannonball never touches the ground. It would continue moving horizontally indefinitely if there were no external factors (like air resistance) affecting its trajectory.

Therefore, the cannonball spends an infinite amount of time in the air.