Question

A ball is thrown horizontally from the top of a building that is \(26.01 \mathrm{~m}\) high. The ball strikes the ground at an angle of \(41.86^{\circ}\) with respect to the horizontal. Neglecting air resistance, with what speed was the ball thrown?

Short answer

Answer: The initial speed of the ball can be calculated using the following equation: \(v_0 = \sqrt{\dfrac{26.01 \mathrm{~m} \cdot s_x}{\tan(41.86^{\circ}) - \dfrac{1}{2}(9.81 \mathrm{~m/s^2})s_x}}\) In this equation, \(s_x\) is the horizontal distance traveled by the ball, which can be determined from the given information. Once the value of \(s_x\) is known, the initial speed, \(v_0\), can be calculated.

Step-by-step solution

Express equations for horizontal and vertical motion

Let's start by expressing equations for horizontal and vertical motion. For horizontal motion, we have: \(s_x = v_0 t\) (1) For vertical motion, we have: \(s_y = h + \dfrac{1}{2}gt^2\) (2) where \(s_x\) is the horizontal distance, \(s_y\) is the vertical distance, \(v_0\) is the initial velocity, \(h\) is the height, and \(g\) is acceleration due to gravity. As the ball is thrown horizontally and the angle at which it strikes the ground is given, we can use the properties of a right triangle to relate horizontal and vertical distances.

Relate the horizontal and vertical distances

Using the angle at which the ball strikes the ground, we can find the relationship between horizontal and vertical distances as follows: \(\tan(41.86^{\circ}) = \dfrac{s_y}{s_x}\) \(=> s_y = s_x \tan(41.86^{\circ})\) (3) Now, we can substitute (3) in equation (2) to eliminate \(s_y\).

Combine the equations and find the time of flight

Substitute \(s_y = s_x \tan(41.86^{\circ})\) in equation (2): \(h = s_x \tan(41.86^{\circ}) - \dfrac{1}{2}gt^2\) (4) Now, we can substitute \(s_x = v_0 t\) in equation (4): \(h = v_0 t \tan(41.86^{\circ}) - \dfrac{1}{2}gt^2\) (5) Let's solve equation (5) for the time of flight \(t\): \(h = v_0 t \tan(41.86^{\circ}) - \dfrac{1}{2}gt^2\) \(=> t = \dfrac{h}{v_0 \tan(41.86^{\circ}) - \dfrac{1}{2}gt}\) Given that \(h = 26.01 \mathrm{~m}\) and \(g = 9.81 \mathrm{~m/s^2}\)

Determine initial speed by solving the equation

We now have \(t\) in terms of \(v_0\). We can simplify the equation further to find initial speed \(v_0\): \(t = \dfrac{h}{v_0 \tan(41.86^{\circ}) - \dfrac{1}{2}gt}\) We can eliminate \(t\) by substituting equation (1) into the equation for \(t\). \(s_x = v_0 t\) \(=> t = \dfrac{s_x}{v_0}\) Now, substitute \(t = \dfrac{s_x}{v_0}\) into our equation for \(t\): \(\dfrac{s_x}{v_0} = \dfrac{h}{v_0 \tan(41.86^{\circ}) - \dfrac{1}{2}gs_x}\) This equation can be solved for the initial speed \(v_0\): \(v_0^2 = \dfrac{h s_x}{\tan(41.86^{\circ}) - \dfrac{1}{2}gs_x}\) Now, we can plug in the given values for \(h\), \(g\), and \(\tan(41.86^{\circ})\): \(v_0^2 = \dfrac{26.01 \mathrm{~m} \cdot s_x}{\tan(41.86^{\circ}) - \dfrac{1}{2}(9.81 \mathrm{~m/s^2})s_x}\) Finally, solving for \(v_0\) will give us the initial speed of the ball: \(v_0 = \sqrt{\dfrac{26.01 \mathrm{~m} \cdot s_x}{\tan(41.86^{\circ}) - \dfrac{1}{2}(9.81 \mathrm{~m/s^2})s_x}}\)

Key concepts

Kinematics Equations

Kinematic equations are the foundations of projectile motion that describe the relationships between displacement, velocity, acceleration, and time. In brief, they come in handy for calculating different parameters of the object's motion when it is either thrown into the air or falling down under the influence of gravity.

For projectile motion, we use two key kinematics equations. The first one, \( s_x = v_0 t \), refers to horizontal displacement where \( s_x \) is the horizontal distance traveled, \( v_0 \) is the initial velocity, and \( t \) is the time of flight. The second, \( s_y = h + \frac{1}{2}gt^2 \), defines vertical displacement where \( s_y \) is the vertical distance fallen, \( h \) is the initial height, and \( g \) is the acceleration due to gravity.

The power of kinematics lies in its ability to break down motion into vertical and horizontal components, making complex calculations for projectile motion possible.

Horizontal Projectile Motion

Horizontal projectile motion occurs when an object is thrown horizontally from a certain height and moves under the influence of gravity. A common misconception is to mix up horizontal motion with motion at an angle; here, the important distinction is that there's no initial vertical velocity component.

The beauty of physics comes alive in simplifying the laws of motion. It tells us that, unless acted upon by another force, an object in horizontal motion will maintain its initial velocity, which is primarily due to the absence of horizontal acceleration. The motion is thus characterized by a constant horizontal velocity and a vertically accelerating movement due to gravity.

This dual behavior of motion defines an arc-like path, where both components - the horizontal motion and the gravitational pull - work independently and simultaneously. When solving problems involving horizontal projectile motion, one must remember to consider both directions separately and then piece the information together to provide a complete picture of the motion.

Angle of Projectile Impact

The angle of projectile impact is a measure of the inclination of the path of a projectile with respect to the horizontal at the point of impact. This angle provides insights into how steep or shallow the path is as the projectile hits the ground.

One of the interesting aspects about the angle of impact is that it is closely tied to the projectile's initial velocity and the height from which it was thrown. The tangent of the angle can be used to link the vertical and horizontal displacements, as shown by the equation \( \tan(\theta) = \frac{s_y}{s_x} \), where \( \theta \) is the angle of impact.

Understanding the angle of impact is essential for predicting the range, the point of impact, and the components of the final velocity of the projectile. It's an important concept that ties together the physics of motion and trigonometry, often resulting in the ability to further understand and analyze the motion of any projectile in real-life scenarios such as sports, ballistics, and even spaceflight.