Question

Suppose you throw a 0.0520 -kg ball with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\) and at an angle of \(30.0^{\circ}\) above the horizontal from a building \(12.0 \mathrm{~m}\) high. a) What will be its kinetic energy when it hits the ground? b) What will be its speed when it hits the ground?

Short answer

Question: Calculate the kinetic energy and speed of the ball when it hits the ground if it is thrown from a height of 12 m with an initial speed of 10 m/s at an angle of 30 degrees above the horizontal. Answer: The kinetic energy of the ball when it hits the ground is approximately 3.29 J, and the speed of the ball when it hits the ground is approximately 11.2 m/s.

Step-by-step solution

Find the time for the ball to hit the ground

First, we need to find the time it takes for the ball to hit the ground. We should focus on the vertical motion and will employ the following equation: $$y(t) = y_0 + v_{0_y}\cdot t - \frac{1}{2}gt^2$$ The vertical displacement \(y(t)\) is \(-12.0\,\mathrm{m}\) (downward), the initial vertical position \(y_0\) is \(0\,\mathrm{m}\), the initial vertical speed \(v_{0_y}\) can be calculated as \(v_0\cdot \sin(30.0^{\circ})\), and \(g\) is \(9.81\,\mathrm{m/s^2}\) (the acceleration due to gravity). We need to solve for \(t\).

Calculate the vertical velocity component

Calculate the initial vertical speed, \(v_{0_y}\), using the given angle and speed: $$v_{0_y} = v_0\cdot \sin(30.0^{\circ})$$ Plug in the values: $$v_{0_y} = 10.0\,\mathrm{m/s} \cdot \sin(30.0^{\circ}) = 5.0\,\mathrm{m/s}$$

Solve for time

Now we can plug in the values into the equation for vertical motion and solve for \(t\): $$-12.0\,\mathrm{m} = 0\,\mathrm{m} + 5.0\,\mathrm{m/s} \cdot t - \frac{1}{2}(9.81\,\mathrm{m/s^2})t^2$$ We can solve this quadratic equation using the quadratic formula, and we have to take the positive solution for time: $$t \approx 1.229\,\mathrm{s}$$

Calculate the horizontal velocity component

Calculate the horizontal velocity component, \(v_{x}\), which remains constant during the motion: $$v_x = v_0\cdot \cos(30.0^{\circ})$$ Plug in the values: $$v_x = 10.0\,\mathrm{m/s} \cdot \cos(30.0^{\circ}) = 8.66\,\mathrm{m/s}$$

Calculate the final vertical velocity component

Calculate the vertical speed, \(v_y\), when the ball hits the ground using the following equation: $$v_y=v_{0_y}-gt$$ Plug in the values: $$v_y = 5.0\,\mathrm{m/s} - (9.81\,\mathrm{m/s^2})\cdot 1.229\,\mathrm{s} = -7.06\,\mathrm{m/s}$$

Calculate the total speed when the ball hits the ground

Now, we can find the total speed \(v\) when the ball hits the ground using the Pythagorean theorem: $$v = \sqrt{v_x^2 + v_y^2}$$ Plug in the values: $$v = \sqrt{(8.66\,\mathrm{m/s})^2 + (-7.06\,\mathrm{m/s})^2} \approx 11.2\,\mathrm{m/s}$$

Calculate the kinetic energy

Now, we can calculate the kinetic energy using the following formula: $$KE = \frac{1}{2}mv^2$$ Plug in the values: $$KE = \frac{1}{2}(0.0520\,\mathrm{kg})(11.2\,\mathrm{m/s})^2\approx3.29\,\mathrm{J}$$ Therefore: a) The kinetic energy of the ball when it hits the ground is approximately \(3.29\,\mathrm{J}\). b) The speed of the ball when it hits the ground is approximately \(11.2\,\mathrm{m/s}\).

Key concepts

Kinetic Energy Calculation

Kinetic energy represents the energy that an object possesses due to its motion. When solving problems involving the kinetic energy of a projectile, like a ball thrown from a building, you calculate it using the formula:

\[\begin{equation}KE = \frac{1}{2}mv^2\right)
where \(KE\) is the kinetic energy, \(m\) is the mass of the object, and \(v\) is the velocity of the object when it hits the ground. It's important to note that kinetic energy is directly proportional to the mass and grows with the square of the velocity; hence, small changes in velocity result in larger changes in kinetic energy. This explains why having an accurate measure of the final velocity is crucial in kinetic energy calculations.

Quadratic Motion Equations

Quadratic motion equations describe the parabolic trajectory of an object under the influence of gravity in projectile motion scenarios. For vertical motion, the standard form is:

\[\begin{equation}y(t) = y_0 + v_{0_y} \times t - \frac{1}{2}gt^2\right)
where \(y(t)\) is the final position, \(y_0\) is the initial position, \(v_{0_y}\) is the initial component of velocity in the vertical direction, \(g\) is the acceleration due to gravity, and \(t\) is the time. The term \(-\frac{1}{2}gt^2\) illustrates the influence of gravity on the object's motion over time, leading to the familiar arc of projectile motion. Solving these equations requires techniques from algebra, particularly when dealing with the inverted parabola that is characteristic of projectiles returning to the ground.

Components of Velocity

When objects are launched at an angle, their velocity can be divided into horizontal (\(v_x\)) and vertical (\(v_y\)) components. These components can be determined using trigonometric relationships:

\[\begin{equation}v_x = v_0 \times \text{cos}(\theta)\right)
\[\begin{equation}v_y = v_0 \times \text{sin}(\theta)\right)
with \(v_0\) being the initial launch speed and \(\theta\) the launch angle above the horizontal. The horizontal component remains constant if we ignore air resistance, since there are no forces acting horizontally. The vertical component, however, changes over time due to gravity. By understanding these components, you can predict both the path of the projectile and its behavior upon impact, as well as solve for other characteristics such as range, maximum height, and flight duration.