A cannon ball is fired at an angle of 30° above horizontal with an initial speed vi = 55 m/s. Assume the cannon ball leaves the cannon at ground level and ignore air resistance.
(a) How long is the cannon ball in the air?
(b) How far from the base of the cannon does the cannon ball hit the ground?
(c) Determine the correct answer. The speed of the stone
is greatest just after it is launched and just before it hits the ground.
is greatest just after it is launched.
is greatest at its maximum height.
is constant throughout its flight.
Vo = 55m/s[30o]
Xo = 55*Cos30 = 47.63 m/s
Yo = 55*sin30 = 27.5 m/s.
a. Y = Yo + g*Tr = 0 @ max ht.
Tr=-Yo/g = -27.5/-9.8 = 2.806 s.=Rise
time.
Tf = Tr = 2.806 s. = Fall time.
Tr+Tf = 2.806 + 2.806 = 5.61 = Time in
air.
b. Dx = Xo*(Tr+Tf) = 47.63m/s * 5.61s =
267.3 m. from base of cannon.
c. The first choice is correct.
To solve this problem, we will break it down into steps:
Step 1: Analyze the initial parameters
Given:
- Initial speed, vi = 55 m/s
- Launch angle, θ = 30°
- Ignoring air resistance
Step 2: Calculate the initial velocity components
The initial velocity can be split into horizontal (vx) and vertical (vy) components using trigonometry:
vx = vi * cos(θ)
vy = vi * sin(θ)
Plugging in the values:
vx = 55 * cos(30°) ≈ 47.57 m/s
vy = 55 * sin(30°) ≈ 27.5 m/s
Step 3: Determine the time of flight (a)
Since the cannonball is in the air for the same amount of time it takes to reach the maximum height and then fall back to the ground, we can calculate the time of flight (t) as:
t = (2 * vy) / g, where g is the acceleration due to gravity (approximately 9.8 m/s²)
Plugging in the values:
t = (2 * 27.5) / 9.8 ≈ 5.61 seconds
Therefore, the cannonball is in the air for approximately 5.61 seconds.
Step 4: Determine the horizontal distance (b)
The horizontal distance traveled (range) can be calculated as:
R = vx * t
Plugging in the values:
R = 47.57 * 5.61 ≈ 267.23 meters
Therefore, the cannonball hits the ground approximately 267.23 meters away from the base of the cannon.
Step 5: Determine the correct answer (c)
The correct answer is: "The speed of the stone is greatest just after it is launched."
To find the answers to these questions, we can use principles of projectile motion. We will use the following equations:
1. The time of flight equation:
T = 2 * (vi * sin(theta)) / g
2. The horizontal displacement equation:
R = vi * cos(theta) * T
3. The maximum height equation:
H = (vi^2 * sin^2(theta)) / (2 * g)
4. The velocity equation at any point during the motion:
v = sqrt((vi*cos(theta))^2 + (vi*sin(theta) - g*t)^2)
Now, let's find the answers to the given questions:
(a) How long is the cannon ball in the air?
To find the time of flight (T), we can use equation 1. Plug in the given values: vi = 55 m/s and theta = 30°, and assume g = 9.8 m/s^2.
T = 2 * (55 * sin(30°)) / 9.8
T ≈ 6.33 seconds
Therefore, the cannonball is in the air for approximately 6.33 seconds.
(b) How far from the base of the cannon does the cannon ball hit the ground?
To find the horizontal displacement (R), we can use equation 2. Plug in the given values: vi = 55 m/s, theta = 30°, and T ≈ 6.33 s.
R = 55 * cos(30°) * 6.33
R ≈ 273.83 meters
Therefore, the cannonball hits the ground approximately 273.83 meters from the base of the cannon.
(c) Determine the correct answer. The speed of the stone:
To determine the correct answer, let's address each option:
- The speed of the stone is greatest just after it is launched and just before it hits the ground.
This option is not correct because the stone experiences vertical acceleration due to gravity, so its speed decreases as it goes up and increases as it comes down.
- The speed of the stone is greatest just after it is launched.
This option is also not correct given the explanation above. The speed is not constant throughout the motion, but it reaches its maximum value somewhere in between.
- The speed of the stone is greatest at its maximum height.
This option is also not correct because at the maximum height, the vertical component of the velocity is zero, and the speed would be maximum at a different point in the motion.
- The speed of the stone is constant throughout its flight.
This option is also not correct as the speed of the stone changes due to gravity, increasing and then decreasing during the flight.
Therefore, none of the given options is correct. The speed of the stone changes throughout its flight due to gravity.