King Arthur's knights fire a cannon from the top of the castle wall. The cannonball is fired at a speed of 42 m/s and at an angle of 33°. A cannonball that was accidentally dropped hits the moat below in 1.6 s.
(a) How far from the castle wall does the cannonball hit the ground?
(b) What is the ball's maximum height above the ground?
Use the dropped cannonball's fall time (1.6 s) to compute the height of the top of the castle.
H = (g/2)t^2 = 12.54 m
(a) In this case, assume they are talking about the fired cannonball, not the dropped one.
Solve this equation for time of flight t:
H = 12.54 -(g/2)t^2 + 42sin33 t = 0
Then use
X = 42cos33*t
b) Ymax = H + (42sin33)^2/(2g)
Note that the castle height H had to be added to the distance that the cannonball actually rises.
Oh, the treacherous tale of King Arthur's knights and their misadventures with cannons! Let's calculate their explosive predicament, shall we?
(a) How far from the castle wall does the cannonball hit the ground? Well, to determine the horizontal distance, we need to find the time it takes for the cannonball to hit the ground. Since we have the initial speed of the cannonball, we can use some trigonometry to split that velocity into horizontal and vertical components.
The horizontal component of the velocity is given by Vx = V * cos(θ), where V is the magnitude of the initial velocity and θ is the angle of elevation. So, Vx = 42 m/s * cos(33°) = 35.26 m/s.
To find the time, we can use the vertical motion equation h = Vyi * t + (1/2) * g * t^2, where h is the vertical distance traveled, Vyi is the initial vertical velocity component, t is the time, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the ball was dropped accidentally, the initial vertical velocity component is 0, and the equation simplifies to h = (1/2) * g * t^2.
Plugging in known values, we have 0 = (1/2) * 9.8 m/s^2 * (1.6 s)^2. Solving for the vertical distance h, we find h = 12.544 m.
Now, knowing the horizontal velocity component (Vx) and the time (t), we can find the horizontal distance d traveled by using the equation d = Vx * t.
Plugging in the values, we have d = 35.26 m/s * 1.6 s = 56.416 m.
So, the cannonball hits the ground approximately 56.416 meters away from the castle wall. Ah, what an explosive result!
(b) Now, let's determine the ball's maximum height above the ground – its crowning glory before gravity claims it. To find the maximum height, we can use the equation Vfy = Vyi + g * t, where Vfy is the final vertical velocity component.
Given that the cannonball was dropped accidentally, Vfy = 0 m/s. We also know that Vyi = V * sin(θ), which becomes V * sin(33°) for our case.
Using this information and rearranging the equation, we get t = - Vyi / g.
Plugging in the values, we have t = - (42 m/s * sin(33°)) / 9.8 m/s^2.
Simplifying, we find t ≈ -0.262 s.
Since we are interested in the absolute value of time, we ignore the negative sign, and the time becomes approximately 0.262 s.
Now, to find the maximum height h_max, we can use the equation h_max = Vyi * t + (1/2) * g * t^2.
Plugging in our values, we have h_max = (42 m/s * sin(33°)) * 0.262 s + (1/2) * 9.8 m/s^2 * (0.262 s)^2.
Calculating, we find h_max ≈ 7.110 m.
Hence, the cannonball's maximum height above the ground is approximately 7.110 meters – a true aerial marvel for a momentarily graceful cannonball.
To solve this problem, we need to use the equations of motion for projectile motion. Let's break down each part of the problem step-by-step.
(a) How far from the castle wall does the cannonball hit the ground?
Step 1: Separate the velocity into horizontal and vertical components.
v₀_x = v₀ * cos(θ)
v₀_y = v₀ * sin(θ)
Given:
v₀ = 42 m/s
θ = 33°
Step 2: Find the time of flight.
We can use the vertical component to find the time it takes for the cannonball to hit the ground.
Since the ball is accidentally dropped, its initial velocity in the y-direction is 0 m/s.
Using the equation:
h = v₀_y * t + 1/2 * g * t²
where h = 0 (because the ball hits the ground), v₀_y = v₀ * sin(θ), and g = 9.8 m/s² (acceleration due to gravity)
We can rewrite the equation as:
0 = v₀ * sin(θ) * t + 1/2 * g * t²
Simplifying, we get:
1/2 * g * t² = - v₀ * sin(θ) * t
Rearranging the equation, we get:
1/2 * g * t² + v₀ * sin(θ) * t = 0
Solving this quadratic equation for t, we get two solutions, but we only need the positive solution for the time of flight.
Let's use the quadratic formula to find t:
t = (-b + √(b² - 4ac)) / (2a)
where a = 1/2 * g, b = v₀ * sin(θ), and c = 0
Calculating the values, we get:
a = 1/2 * 9.8 = 4.9
b = 42 * sin(33°) ≈ 22.06
c = 0
Using the quadratic formula:
t = (-22.06 + √(22.06² - 4 * 4.9 * 0)) / (2 * 4.9)
t ≈ 2.028 s
The time of flight is approximately 2.028 seconds.
Step 3: Find the horizontal distance traveled.
Now that we know the time of flight, we can find the horizontal distance traveled by using:
d = v₀_x * t
where v₀_x = v₀ * cos(θ) and t = time of flight
Using the given values:
v₀_x = 42 * cos(33°) ≈ 35.23
t ≈ 2.028
Calculating the values, we get:
d ≈ 35.23 * 2.028
d ≈ 71.41 m
The cannonball hits the ground approximately 71.41 meters from the castle wall.
(b) What is the ball's maximum height above the ground?
Step 1: Calculate the time it takes to reach the highest point.
Since the ball is launched at an angle, we need to find the time it takes to reach the highest point of its trajectory.
We can use the equation:
v_fy = v₀_y - g * t
where v_fy = 0 (at the highest point), v₀_y = v₀ * sin(θ), and g = 9.8 m/s²
At the highest point, the vertical velocity becomes zero.
0 = v₀ * sin(θ) - g * t
Simplifying, we get:
v₀ * sin(θ) = g * t
Solving for t, we get:
t = v₀ * sin(θ) / g
Using the given values:
v₀ = 42 m/s
θ = 33°
g = 9.8 m/s²
Calculating the values, we get:
t = 42 * sin(33°) / 9.8
t ≈ 1.68 s
It takes approximately 1.68 seconds for the cannonball to reach the highest point.
Step 2: Calculate the maximum height.
We can use the vertical component of the equation of motion to find the maximum height.
Using the equation:
h = v₀_y * t + 1/2 * g * t²
where v₀_y = v₀ * sin(θ), t = time to reach the highest point, and g = 9.8 m/s²
Substituting the values we know:
v₀_y = 42 * sin(33°)
t ≈ 1.68
Calculating the values, we get:
h ≈ (42 * sin(33°)) * 1.68 + 1/2 * 9.8 * (1.68)²
h ≈ 49.125 m
The maximum height above the ground is approximately 49.125 meters.
To solve this problem, we can use the equations of motion for projectiles. We will consider the vertical and horizontal components of the cannonball's motion separately.
(a) To find how far from the castle wall the cannonball hits the ground, we need to determine the horizontal distance traveled by the cannonball.
First, let's find the time it takes for the cannonball to hit the ground. We can use the equation for vertical displacement:
y = V₀y * t - (1/2) * g * t²
where:
- y is the vertical displacement (which is the height of the moat),
- V₀y is the initial vertical velocity (which is given as 0 for an object dropped from rest),
- g is the acceleration due to gravity, which is approximately 9.8 m/s²,
- t is the time taken.
Since the cannonball takes 1.6 s to hit the moat, we can substitute these values into the equation and solve for y:
0 = 0 + (1/2) * 9.8 * (1.6)²
This equation simplifies to:
0 = 12.544 - 0.5 * 9.8 * (1.6)²
0 = 12.544 - 0.5 * 9.8 * 2.56
0 = 12.544 - 12.544
0 = 0
Since 0 = 0 is a true statement, it means the cannonball indeed hits the moat after 1.6 s.
Now let's solve for the horizontal distance traveled by the cannonball. We can use the equation for horizontal displacement:
x = V₀x * t
where:
- x is the horizontal displacement,
- V₀x is the initial horizontal velocity (which can be found using the initial speed and angle),
- t is the time taken.
The initial horizontal velocity can be found using the initial speed and angle:
V₀x = V₀ * cosθ
where:
- V₀ is the initial speed (42 m/s),
- θ is the angle (33°).
Substituting the values into the equation, we can find V₀x:
V₀x = 42 * cos(33°)
V₀x ≈ 35.246 m/s
Finally, we can substitute the values of V₀x and t into the equation for horizontal displacement to find the distance from the castle wall:
x = (35.246) * (1.6)
x ≈ 56.394 m
Therefore, the cannonball hits the ground approximately 56.394 meters from the castle wall.
(b) To find the maximum height above the ground, we can use the equation for vertical displacement once again. This time, we'll find the time it takes for the ball to reach its maximum height.
Since the vertical velocity at the maximum height is 0, we can use the equation:
Vf = Vi + g * t
where:
- Vf is the final vertical velocity (0 m/s),
- Vi is the initial vertical velocity (which can be found using the initial speed and angle),
- g is the acceleration due to gravity (approximately 9.8 m/s²),
- t is the time taken.
Substituting the values, we can solve for t:
0 = V₀y + (9.8) * t
Since the initial vertical velocity is given by V₀ * sinθ, we have:
0 = (V₀ * sinθ) + (9.8) * t
Substituting the values of V₀ and θ, we can solve for t:
0 = (42 * sin(33°)) + (9.8) * t
0 = 21.6 + (9.8) * t
-21.6 = (9.8) * t
t = -21.6 / 9.8
t ≈ -2.204 s
Since time cannot be negative in this context, we discard the negative value. Thus, the time at the maximum height is approximately 0.2 seconds.
Now, let's find the maximum height above the ground. We can use the equation for vertical displacement:
y = V₀y * t - (1/2) * g * t²
Substituting the values, we have:
y = (0 * 0.2) - (1/2) * 9.8 * (0.2)²
y = 0 - (1/2) * 9.8 * 0.04
y = 0 - 0.784
y ≈ -0.784 m
The negative sign indicates that the height is below the initial position. Therefore, the ball's maximum height is approximately 0.784 meters below the initial position (ground level).
So, the ball does not reach above the initial height.