A cannonball is catapulted toward a castle. The cannonball's velocity when it leaves the catapult is 39 m/s at an angle of 40° with respect to the horizontal and the cannonball is 7.0 m above the ground at this time.
(a) What is the maximum height above the ground reached by the cannonball?
1 m
(b) Assuming the cannonball makes it over the castle walls and lands back down on the ground, at what horizontal distance from its release point will it land?
2 m
(c) What are the x- and y-components of the cannonball's velocity just before it lands? The y-axis points up.
3 m/s (x-component)
4 m/s (y-component)
Vo = 39m/s @ 40 Deg.
Xo = 39*cos40 = 29.9 m/s.
Yo = 39*sin40 = 25.1 m/s.
a. hmax = ho + (Yf^2-Yo^2)/2g,
hmax = 7 + (0-(25.1)^2) / -19.8=39.1 m.
b. Dx = Vo^2*sin(2A)/g, 2A = 80 Deg.
c. Xo = 39*cos40 = 29.9 m/s.
Yf^2 = Vo^2 + 2g*hmax,
Yf^2 = 0 + 19.6*39.1 = 766.36,
Yf = 27.7 m/s.
To find the answers to these questions, we can use the equations of motion to analyze the motion of the cannonball. Let's break down the problem step by step.
Step 1: Find the initial vertical velocity (Vy) and horizontal velocity (Vx) components of the cannonball's velocity.
Given:
Initial velocity (V) = 39 m/s
Angle (θ) = 40°
Using the given information, we can find the vertical and horizontal components of the velocity using trigonometry.
Vy = V * sin(θ)
= 39 m/s * sin(40°)
≈ 25 m/s
Vx = V * cos(θ)
= 39 m/s * cos(40°)
≈ 29.8 m/s
Step 2: Find the time it takes for the cannonball to reach maximum height.
To find the time taken to reach the maximum height (t), we can use the equation:
Vy = Vy_initial + (g * t)
0 = 25 m/s + (-9.8 m/s^2) * t
Solving for t:
t = -25 m/s / -9.8 m/s^2
t ≈ 2.55 s
Step 3: Find the maximum height reached by the cannonball.
Using the time calculated in the previous step, we can find the maximum height (h) reached by the cannonball using the equation:
h = h_initial + (Vy_initial * t) + (0.5 * g * t^2)
= 7.0 m + (25 m/s * 2.55 s) + (0.5 * -9.8 m/s^2 * (2.55 s)^2)
≈ 39.2 m
Therefore, the maximum height above the ground reached by the cannonball is approximately 39.2 meters.
Step 4: Find the time taken for the cannonball to reach the ground.
To find the total time taken for the cannonball to reach the ground, we can use the equation:
h = h_initial + (Vy_initial * t) + (0.5 * g * t^2)
0 = 39.2 m + (25 m/s * t) + (0.5 * -9.8 m/s^2 * t^2)
Since we already found the time taken to reach maximum height (t), we can use it as an initial guess to find the final time using numerical methods (e.g., quadratic formula, graphing, etc.). However, without more information or an explicit equation relating time and position, it is not possible to provide a specific value for the time taken to reach the ground in this case.
Step 5: Find the horizontal distance traveled by the cannonball.
To find the horizontal distance traveled by the cannonball, we can use the equation:
x = Vx * t
Using the initial horizontal velocity (Vx) and the time t calculated in Step 2, we can find the horizontal distance traveled by the cannonball:
x = 29.8 m/s * 2.55 s
≈ 76.19 m
Therefore, the cannonball will land at a horizontal distance of approximately 76.19 meters from its release point.
Step 6: Find the x- and y-components of the cannonball's velocity just before it lands.
Since air resistance is not considered, the horizontal velocity (Vx) of the cannonball remains constant throughout its trajectory. Therefore, the x-component of the cannonball's velocity just before it lands will be the same as the initial horizontal velocity:
x-component = Vx = 29.8 m/s
The y-component of the cannonball's velocity just before landing can be found using the equation:
Vy = Vy_initial + (g * t)
Substituting the known values:
Vy = 25 m/s + (-9.8 m/s^2) * t
Since we don't have the specific value for the time taken to reach the ground, we cannot determine the exact value for the y-component of velocity just before the cannonball lands. Therefore, without additional information, we cannot provide a specific value for the y-component of the velocity just before it lands.
To solve this problem, we can use the principles of projectile motion. We'll break down the motion of the cannonball into its horizontal and vertical components.
(a) To find the maximum height above the ground reached by the cannonball, we need to determine the vertical component of the initial velocity. Given that the initial velocity of the cannonball is 39 m/s at an angle of 40° with respect to the horizontal, we can find the vertical component using trigonometry.
The vertical component of the initial velocity (Vy) is given by Vy = V * sin(θ), where V is the magnitude of the initial velocity (39 m/s) and θ is the launch angle (40°). Substituting the values, we have:
Vy = 39 m/s * sin(40°)
Vy ≈ 25.03 m/s
Next, we can use the kinematic equation to find the maximum height (h) reached by the projectile. The equation is:
Vy^2 = Vo^2 + 2 * a * h
Since the launch angle is 40° and the cannonball is subjected to only the acceleration due to gravity (-9.8 m/s^2) in the vertical direction, we can rewrite the equation as:
(25.03 m/s)^2 = 0 + 2 * (-9.8 m/s^2) * h
Simplifying the equation, we have:
625.3 m^2/s^2 = -19.6 m/s^2 * h
Solving for h:
h = 625.3 m^2/s^2 / (-19.6 m/s^2)
h ≈ -31.89 m
The negative sign indicates that the cannonball is below the initial launch point at this height calculation. Since it cannot be below ground, we ignore the negative sign and take the magnitude of the height:
Maximum height (h) ≈ 31.89 m
Therefore, the maximum height above the ground reached by the cannonball is approximately 31.89 meters.
(b) To find the horizontal distance from the release point at which the cannonball will land, we need to determine the total time of flight. We can use the vertical component of the initial velocity (25.03 m/s) to calculate the total time (t).
In projectile motion, the time taken for a projectile to reach maximum height and then return to the same height is the same. Hence, we'll double the time it takes to reach the maximum height.
The time taken to reach the maximum height (t1) is given by:
t1 = Vy / a
t1 = 25.03 m/s / 9.8 m/s^2 ≈ 2.55 s
The total time of flight (t) is twice the time taken to reach the maximum height:
t = 2 * t1
t = 2 * 2.55 s ≈ 5.10 s
Now, we can find the horizontal distance using the equation:
dx = Vx * t
The horizontal component of the initial velocity (Vx) is given by Vx = V * cos(θ), where V is the magnitude of the initial velocity (39 m/s) and θ is the launch angle (40°). Substituting the values, we have:
Vx = 39 m/s * cos(40°)
Vx ≈ 29.80 m/s
Calculating the horizontal distance:
dx = Vx * t
dx = 29.80 m/s * 5.10 s
dx ≈ 152.18 m
Therefore, the cannonball will land approximately 152.18 meters away from its release point.
(c) To find the x- and y-components of the cannonball's velocity just before it lands, we can use the following relationships:
Vx = V * cos(θ)
Vy = V * sin(θ)
Given that V is the magnitude of the initial velocity (39 m/s) and θ is the launch angle (40°), we can substitute the values:
Vx = 39 m/s * cos(40°)
Vx ≈ 29.80 m/s
Vy = 39 m/s * sin(40°)
Vy ≈ 25.03 m/s
Therefore, just before it lands, the x-component of the cannonball's velocity is approximately 29.80 m/s, and the y-component is approximately 25.03 m/s.