Question

At a time of \(10.0 \mathrm{~s}\) after being fired, a cannonball strikes a point \(500 . \mathrm{m}\) horizontally from and \(100 . \mathrm{m}\) vertically above the point of launch. a) With what initial velocity was the cannonball launched? b) What maximum height was attained by the ball? c) What is the magnitude and direction of the ball's velocity just before it strikes the given point?

Short answer

The initial velocity of the cannonball is 53.7 m/s. The maximum height attained by the cannonball is 119.4m. The magnitude of the cannonball's velocity just before it strikes the given point is 125.9 m/s, and the direction is 66.7° below the horizontal.

Step-by-step solution

Determine the initial vertical velocity (Vy) using the given distance and time

We are given that the vertical distance above the point of launch is 100m. Let's use the following vertical motion equation and solve for the initial vertical velocity: \(y = V_{y0}t - \frac{1}{2}gt^2\) where \(y = 100m\) \(V_{y0}\)= initial vertical velocity, \(t = 10s\) (time) \(g = 9.81 m/s^2\) (acceleration due to gravity) \(100 = V_{y0}(10) -\frac{1}{2}(9.81)(100)\) Now, solve for \(V_{y0}\): \(V_{y0} = -19.3 m/s\) (Note: The negative sign indicates that the initial vertical velocity is upward)

Determine the initial horizontal velocity (Vx) using the horizontal distance and time

We can use the horizontal motion equation to find the initial horizontal velocity Vx: \(x = V_{x0}t\) where \(x = 500m\) \(V_{x0}\) = initial horizontal velocity, \(t = 10s\) (time) \(500 = V_{x0}(10)\) Now, solve for \(V_{x0}\): \(V_{x0} = 50 m/s\)

Calculate the initial velocity of the cannonball

Now we'll combine the initial vertical and horizontal velocities using the Pythagorean theorem: \(V_0 = \sqrt{V_{x0}^2 + V_{y0}^2}\) \(V_0 = \sqrt{(50)^2 + (-19.3)^2}\) \(V_0 = 53.7m/s\) So, the initial velocity of the cannonball is \(53.7 m/s\).

Calculate the maximum height attained by the cannonball

To find the maximum height, we first need to find the time when the vertical velocity becomes zero. We can use the following equation: \(V_y = V_{y0} - gt\) At the maximum height, \(V_y = 0\). So, solving for the time, \(t_1 = \frac{V_{y0}}{g} = \frac{-19.3}{9.81} = 1.97s\) Now let's find the maximum height (\(h\)), using vertical motion equation: \(h = y + V_{y0}t_1 - \frac{1}{2}gt_1^2\) \(h = 100 - 19.3(1.97) - \frac{1}{2}(9.81)(1.97)^2\) \(h = 119.4m\) So, the maximum height attained by the cannonball is \(119.4m\).

Calculate the magnitude and direction of the ball's velocity just before it strikes the given point

In order to calculate the final vertical velocity, \(V_{yf}\), we use the same equation as in step 1: \(V_{yf} = V_{y0} - gt\) \(V_{yf} = -19.3 - 9.81(10) = -117.3m/s\) The horizontal velocity does not change throughout the motion since there is no air resistance, so \(V_{xf} = 50m/s\) The magnitude of the final velocity can be found using the Pythagorean theorem: \(V_f = \sqrt{V_{xf}^2 + V_{yf}^2}\) \(V_f = \sqrt{(50)^2 + (-117.3)^2}\) \(V_f = 125.9m/s\) To find the direction of the final velocity, we use the inverse tangent function: \(\theta = \tan^{-1}(\frac{V_{yf}}{V_{xf}})\) \(\theta = \tan^{-1}(\frac{-117.3}{50})\) \(\theta = -66.7^{\circ}\) (Note: Negative angle refers to the direction below the horizontal) So, the magnitude and direction of the ball's velocity just before it strikes the given point are \(125.9m/s\) and \(66.7^{\circ}\) below the horizontal.

Key concepts

Initial Velocity

Understanding the initial velocity of an object in projectile motion is crucial because it sets the course for the entirety of the object's trajectory. Initial velocity, often symbolized as V₀, is the speed at which an object is launched into motion. It comprises two components: horizontal velocity (V_x0) and vertical velocity (V_y0). The horizontal component affects how far the object will travel, while the vertical component determines how high it will go. To calculate initial velocity, one can use the Pythagorean theorem, combining these two components. For instance, if a cannonball is launched with a horizontal velocity of 50 m/s and a vertical velocity of -19.3 m/s, the initial velocity is found using the equation V₀ = \(\.\sqrt{V_{x0}^2 + V_{y0}^2}\).

Maximum Height

The maximum height of a projectile in motion occurs at the peak of its trajectory, where vertical velocity momentarily becomes zero. To find this point, we can use kinematic equations. In particular, the time (t₁) it takes for the vertical velocity to reach zero can indicate when the projectile is at its peak. Then, the maximum height (h) can be calculated using the initial vertical velocity and the time it takes to reach that peak. For example, the formula h = y + V_y0t₁ - \(\frac{1}{2}g\)t₁^2 gives us the maximum height of a projectile. The trajectory's apex is a critical point in defining the path and potential impact of the object in motion.

Velocity Components

In projectile motion, understanding an object's velocity entails dividing it into two separate components: horizontal (V_x) and vertical (V_y). The horizontal velocity remains constant due to the absence of horizontal forces (when ignoring air resistance), while the vertical velocity changes due to gravity. The components are essential for analyzing the motion separately in each direction. Using trigonometry, we can decompose the initial velocity into these components, and we can also analyze the final velocity, as with the cannonball that has a final vertical velocity of -117.3 m/s and a horizontal velocity of 50 m/s just before impact.

Kinematic Equations

Kinematic equations allow for the prediction of various aspects of an object's motion, like displacement, velocity, and time, under constant acceleration. They are essential tools in understanding projectile motion physics. These equations do not consider the mass of the object and assume acceleration is constant, typically the acceleration due to gravity (g) for vertically moving objects. The kinematic equations are versatile and can solve for the unknown values when certain information, such as initial velocity, time, or distance, is provided. A common equation used to find the vertical displacement is y = V_y0t - \(\frac{1}{2}g\)t^2. To determine the various motion parameters of a projectile, it is often necessary to first break down its velocity into its components and then use these equations accordingly.