Answer:

We can use the kinematic equations of motion to solve this problem.

First, we can split the initial velocity into horizontal and vertical components:

- The horizontal component is v₀x = v₀cosθ = 60cos(30°) ≈ 51.96 m/s

- The vertical component is v₀y = v₀sinθ = 60sin(30°) ≈ 30 m/s

Next, we can use the equation for the vertical displacement of an object under constant acceleration:

Δy = v₀yt + 1/2at²

where Δy = -50 m (since the cannonball is moving downwards), v₀y = 30 m/s, and a = -9.81 m/s² (the acceleration due to gravity). Solving for t, we get:

-50 m = 30 m/s * t + 1/2 * (-9.81 m/s²) * t²

-50 m = 30t - 4.905t²

4.905t² - 30t - 50 = 0

Using the quadratic formula, we get:

t = [30 ± sqrt(30² - 4*4.905*(-50))] / (2*4.905)

t ≈ 6.56 s (ignoring the negative root, since time cannot be negative)

So the cannonball takes approximately 6.56 seconds to hit the ground.

To find its velocity when it hits the ground, we can use the equation for the horizontal displacement of an object under constant velocity:

Δx = v₀xt

where Δx = v * t (the distance traveled before hitting the ground), v₀x = 51.96 m/s, and t = 6.56 s. Solving for v, we get:

v = Δx / t = (51.96 m/s) * (6.56 s) ≈ 340.2 m/s

So the velocity of the cannonball when it hits the ground is approximately 340.2 m/s.

Explanation:

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