Question

In Exercises 25 and 26 , use the vertical motion model \(\boldsymbol{h}=-\mathbf{1 6 t}^{2}+\boldsymbol{v t}+\boldsymbol{s}(\mathbf{p} . \mathbf{5 3 5})\) and the following information. You and a friend are playing basketball. You can jump with an initial velocity of 12 feet per second. You need to jump 2.2 feet to dunk a basketball. Your friend can jump with an initial velocity of 14 feet per second. Your friend needs to jump 3.4 feet to dunk a basketball. Can you dunk the ball? Can your friend? Justify your answers.

Short answer

You can dunk a basketball as your maximum jump height is 2.25 feet, which is more than the required 2.2 feet. Similarly, your friend can also dunk the basketball because their maximum jump height is 3.0625 feet, which is also more than the required 3.4 feet.

Step-by-step solution

Substitute your data into the model

First, substitute your initial velocity (12 feet per second) and initial height (0 being on ground level) into the motion model equation, resulting in \( h = -16t^{2} + 12t + 0 \).

Determine the time of the peak of your jump

The time at which the peak of the jump is reached can be determined by setting the equation derivative to zero i.e., \( -32t + 12 = 0 \). Solving for \( t \) gives \( t = 0.375 \) seconds.

Compute your maximum jump height

Substitute your peak time into your height equation, i.e., \( h = -16(0.375)^{2} + 12(0.375) + 0 \) and solve to get \( h = 2.25 \) feet.

Substitute your friend's data into the model

Now, substitute your friend's initial velocity (14 feet per second) and initial height (0 being on ground level) into the motion model equation, resulting in the new equation \( h = -16t^{2} + 14t + 0 \).

Determine the time of the peak of your friend's jump

The time at which the peak of the jump is reached can be determined by setting the equation derivative to zero i.e., \( -32t + 14 = 0 \). Solving for \( t \) gives \( t = 0.4375 \) seconds.

Compute your friend's maximum jump height

Substitute your friend's peak time into the height equation, i.e., \( h = -16(0.4375)^{2} + 14(0.4375) + 0 \). The solution is \( h = 3.0625 \) feet.

Key concepts

Quadratic Equations

Quadratic equations are mathematical expressions of the form \( ax^2 + bx + c = 0 \), where \( a \) is the coefficient of the squared term, \( b \) is the coefficient of the linear term, and \( c \) is the constant. They are pivotal in various disciplines, including physics, particularly in analyzing the motion of objects.

The vertical motion of a projectile, like a basketball shot or a high jump, can be modeled by a quadratic equation. In these equations, the variable \( t \) usually stands for time, and the equation itself represents the height (\( h \) ) of the object at any given time. The coefficients in the equation correspond to the physical factors of motion: \( -16t^2 \) represents the downward force of gravity (assuming feet per second squared as the unit), and the \( vt \) term represents the initial velocity of the object in the direction of the motion (upwards, in this case).

To solve these equations and find the height at any time, or the time it takes to reach a certain height, one can use various methods such as factoring, completing the square, or the quadratic formula. Understanding how to manipulate these equations is essential for analyzing the motion of projectiles.

Initial Velocity

Initial velocity refers to the speed at which an object starts its motion. In the context of vertical motion, it signifies the upward speed of the object at the moment it leaves the ground. The initial velocity is a crucial factor because it sets the stage for the entire trajectory of the projectile.

In our exercise, the initial velocity appears in the equation as the \( vt \) term. For you, it's 12 feet per second, and for your friend, it's 14 feet per second. A higher initial velocity typically results in a higher and longer trajectory, provided that other conditions like the angle of release and air resistance remain constant.

Mathematically, a change in initial velocity will alter the quadratic equation, subsequently affecting the calculation of the maximum height reached by the object (in this case, the heights of the jumps). The initial velocities contribute significantly to the height reached before gravity pulls the object back down.

Maximum Height of Projectile

The maximum height of a projectile is the highest vertical position the projectile reaches in its trajectory. It's determined by both the force of gravity and the projectile's initial velocity. In a vertical motion model like \( h = -16t^2 + vt + s \), the maximum height can be found by calculating the vertex of the parabola represented by the quadratic equation, which occurs at the peak of the object's trajectory.

As shown in the step-by-step solution, we can find the time at which the peak occurs by setting the derivative of the height equation to zero and then solving for \( t \) to find when the upward velocity is momentarily zero before gravity overtakes the initial velocity and starts pulling the projectile down. The height at this time will give us the maximum height of the projectile.

For you, with an initial velocity of 12 feet per second, the maximum height reached is 2.25 feet, while your friend, with an initial velocity of 14 feet per second, reaches 3.0625 feet. This information determines whether you and your friend can dunk a basketball, as you would need to overcome the respective vertical distances needed to dunk.