A cannon mounted on a pirate ship fires a cannonball at 125 m/s horizontally, at a height of 17.5 m above the ocean surface. Ignore air resistance. (a) How much time elapses until it splashes into the water? (b) How far from the ship does it land?
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To determine the time it takes for the cannonball to splash into the water, we can use the equation of motion in the vertical direction. The initial vertical velocity is zero, as the cannonball is fired horizontally. The height above the water's surface is 17.5 m. The acceleration due to gravity, denoted as "g," is approximately 9.8 m/s^2. We will consider the upward direction as positive and the downward direction as negative.
(a) Time elapsed until the cannonball splashes into the water:
Using the equation of motion in the vertical direction:
Δy = v₀t + (1/2)gt²,
where Δy is the displacement in the vertical direction, v₀ is the initial vertical velocity (zero in this case), t is the time, and g is the acceleration due to gravity.
Since the cannonball falls downward, the displacement Δy is equal to the negative height above the water's surface, -17.5 m. The equation becomes:
-17.5 = 0 + (1/2) * 9.8 * t².
Rearranging the equation to solve for t:
-17.5 = 4.9t²,
t² = -17.5 / 4.9,
t² ≈ 3.5714.
Taking the square root of both sides:
t ≈ √3.5714,
t ≈ 1.888 s.
Therefore, it takes approximately 1.888 seconds for the cannonball to splash into the water.
(b) Horizontal distance traveled by the cannonball:
Since there is no horizontal acceleration, the cannonball maintains a constant horizontal velocity throughout its flight.
To find the horizontal distance traveled, we can multiply the horizontal velocity (125 m/s) by the time t.
Distance = Velocity × Time,
Distance = 125 m/s × 1.888 s,
Distance ≈ 236.0 m.
Hence, the cannonball lands approximately 236.0 meters from the ship.