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?
(a) Since there is no initial vertical velocity component,
H = (1/2) g t^2
(The height above water is H.)
t = sqrt(2H/g) = 0.529 s
(b) 0.529 s * 125 m/s = 66.1 meters
The cannonball would have travelled much farther if it had been pointed upward.
To answer this question, we can use the basic principles of projectile motion. Let's break it down into two parts:
(a) How much time elapses until the cannonball splashes into the water?
To find the time taken by the cannonball to reach the water, we need to consider the vertical motion of the cannonball. We'll use the equation:
d = v₀t + 0.5at²
where
d = displacement (17.5 m, since the initial height is given)
v₀ = initial vertical velocity (0 m/s because the cannonball was launched horizontally)
t = time taken (what we're trying to find)
a = acceleration due to gravity (-9.8 m/s² in the opposite direction of motion)
Plugging in the values, we get:
17.5 m = 0 * t + 0.5 * (-9.8 m/s²) * t²
Simplifying the equation, we have:
-4.9t² = 17.5
To find t, we can rearrange the equation as follows:
t² = 17.5 / -4.9
t² ≈ -3.57
Now, we can take the square root of both sides:
t ≈ √(-3.57)
However, in this case, the square root of a negative number doesn't have a real solution. It indicates that the cannonball will never reach the surface of the water unless there's an external force acting on it (e.g., wind, propulsion).
(b) How far from the ship does the cannonball land?
Since we know the horizontal velocity of the cannonball is 125 m/s and the time is not known, we can calculate the horizontal displacement using the equation:
d = v₀t
where
d = horizontal displacement (what we're trying to find)
v₀ = initial horizontal velocity (125 m/s)
t = time taken (unknown)
Rearranging the equation, we get:
t = d / v₀
Since time can be written as the time taken by the cannonball to reach the water, we can use d = v₀ₓt. Here, v₀ₓ is the horizontal component of the initial velocity, which can be calculated using v₀cosθ, where θ is the angle made by the cannonball with the horizontal (0 degrees in this case).
So, v₀ₓ = v₀ * cos(0)
v₀ₓ = 125 m/s * 1
v₀ₓ = 125 m/s
Hence, d = v₀ₓt becomes d = 125 m/s * t.
Since we already determined that it will never reach the water, the answer is that the cannonball will not land anywhere.