A canon mounted on a pirate ship fires a cannonball at 125 m/s horizontally, at a height of 17.5m 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) Ahoy there! Let me grab my trusty parrot, Polly Math, to help calculate this. Polly, our pirate ship cannonball friend is launched horizontally at 125 m/s. Given that the height of the cannon is 17.5m above the ocean surface, we can use the good old kinematic equation:
h = ut + ½gt²
Where:
h = height (17.5m)
u = initial velocity (0 m/s - since we're considering only the horizontal component)
g = acceleration due to gravity (9.8 m/s²)
t = time elapsed
Since we're aiming for the splash in the water, we want to find the time it takes for the cannonball to reach the ground, which means the final height (h) becomes zero. So we rearrange the equation:
0 = 0.5 * 9.8 * t²
Simplifying this equation, we find that t² = 0. Therefore, the time taken for the cannonball to splash into the water is... instantaneous! It's a quick splash indeed!
(b) Now, to calculate how far from the ship the cannonball lands. We know the horizontal velocity is 125 m/s and the time it takes to splash (t) is 0 seconds. To find the distance traveled (d), we can use another simple equation:
d = vt
Let's put in the values:
d = 125 * 0
Oh dear, multiplication by zero isn't the brightest idea, even for pirates! It seems our cannonball hasn't traveled any horizontal distance. So it lands at exactly the same spot it was fired from. Perhaps we need a pirate with better aim or a more adventurous cannonball next time!
Avast, matey! Keep those cannons firing, but be careful with the maths!
To find the solution, we can use the kinematic equations of motion. Let's break down each part of the problem.
(a) How much time elapses until the cannonball splashes into the water?
The cannonball is fired horizontally, which means its initial vertical velocity (Vy) is 0 m/s. The only force acting on the cannonball in the vertical direction will be gravity.
We can use the equation:
Δy = Vyi * t + (1/2) * g * t^2
Where:
Δy = vertical displacement (17.5 m)
Vyi = initial vertical velocity (0 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
t = time
Plugging in the given values, the equation becomes:
17.5 = 0 + (1/2) * (-9.8) * t^2
Simplifying further, we have:
17.5 = -4.9 * t^2
To solve for t, we can rearrange the equation:
t^2 = 17.5 / -4.9
Taking the square root of both sides:
t = √(17.5 / -4.9)
Calculating this value:
t ≈ 1.64 seconds
Therefore, it takes approximately 1.64 seconds for the cannonball to splash into the water.
(b) How far from the ship does it land?
Since the cannonball is fired horizontally, its horizontal velocity (Vx) remains constant at 125 m/s throughout its motion. Therefore, we can calculate the horizontal distance using the equation:
Δx = Vxi * t
Where:
Δx = horizontal distance traveled
Vxi = initial horizontal velocity (125 m/s)
t = time
Plugging in the given values, the equation becomes:
Δx = 125 * 1.64
Calculating this value:
Δx ≈ 205 meters
Therefore, the cannonball lands approximately 205 meters away from the ship.
To find the answers to these questions, we can use the equations of motion under constant acceleration. In this case, the horizontal motion of the cannonball is not affected by gravity since the motion is perpendicular to the vertical direction. However, the vertical motion of the cannonball is influenced by gravity.
Let's break down the problem into two parts: the horizontal and vertical motions.
(a) To find the time it takes for the cannonball to splash into the water, we will focus on the vertical motion. We need to determine the time it takes for the cannonball to fall from a height of 17.5m.
The equation we can use for vertical motion is:
y = y0 + v0y * t - 0.5 * g * t^2
where y is the final vertical position (0m, since it splashes into the water), y0 is the initial vertical position (17.5m), v0y is the initial vertical velocity (0m/s if we are assuming the cannonball is just fired horizontally), t is the time, and g is the acceleration due to gravity (-9.8m/s^2).
Setting y = 0m and rearranging the equation, we have:
0 = 17.5 + 0 * t - 0.5 * 9.8 * t^2
Simplifying the equation, we get:
-4.9 * t^2 = 17.5
Dividing both sides by -4.9, we get:
t^2 = -17.5 / -4.9
t^2 ≈ 3.57
Taking the square root of both sides, we find:
t ≈ sqrt(3.57)
t ≈ 1.89 seconds
Therefore, it takes approximately 1.89 seconds for the cannonball to splash into the water.
(b) To find the horizontal distance the cannonball lands from the ship, we can use the horizontal velocity and the time it takes to splash into the water.
The equation we can use for horizontal motion is:
x = x0 + v0x * t
where x is the horizontal distance, x0 is the initial horizontal position (0m), v0x is the initial horizontal velocity (125m/s), and t is the time we found in part (a) (1.89s).
Plugging in the values, we have:
x = 0 + 125 * 1.89
Simplifying the equation, we get:
x ≈ 236.25m
Therefore, the cannonball lands approximately 236.25 meters from the ship.
a. Vo*t + 0.5g*t^2 = 17.5 m.
0 + 4.9t^2 = 17.5
t^2 = 3.57
Tf = 1.89 s. = Fall time.
b. D = Vo * Tf = 125m/s * 1.89s = 236 m.