Unit 8 Quadratic Equations Homework 14 Projectile Motion Answers

Unit 8 QuadraticEquations Homework 14 Projectile Motion Answers

Projectile motion is a classic application of quadratic equations, and mastering it in Unit 8 helps students see how algebra describes real‑world physics. Homework 14 typically asks learners to model the height of an object launched into the air, find its maximum height, determine the time it hits the ground, and interpret the vertex and roots of the resulting quadratic function. Below is a detailed walk‑through of the concepts, step‑by‑step solutions to typical problems, and practical tips to avoid common errors.

Introduction

When a ball, rocket, or any object is thrown near Earth’s surface, its vertical motion is influenced only by gravity (ignoring air resistance). The height (h) as a function of time (t) follows a quadratic equation of the form

[ h(t)= -\frac{1}{2}gt^{2}+v_{0}t+h_{0}, ]

where * (g) ≈ 9.8 m/s² (acceleration due to gravity),

• (v_{0}) is the initial vertical velocity, and
• (h_{0}) is the launch height.

Because the coefficient of (t^{2}) is negative, the graph is a downward‑opening parabola. The vertex gives the maximum height, and the (t)-intercepts (roots) indicate when the object is at ground level. Unit 8 Quadratic Equations Homework 14 Projectile Motion Answers therefore require students to:

1. Write the correct quadratic model from a word problem.
2. Identify the vertex (time of peak height and maximum height).
3. Solve for the roots (time of launch and time of impact).
4. Interpret the results in context.

Understanding the Quadratic Model

Deriving the Equation

Starting from the kinematic formula for vertical displacement:

[ \Delta y = v_{0}t + \frac{1}{2}at^{2}, ]

set (a = -g) (gravity acts downward) and add the initial height (h_{0}):

[ h(t) = h_{0} + v_{0}t - \frac{1}{2}gt^{2}. ]

Re‑arranging gives the standard quadratic form

[ h(t) = -\frac{1}{2}gt^{2} + v_{0}t + h_{0}. ]

Key Features

Step‑by‑Step Solution to Typical Homework Problems

Below are three representative problems that often appear in Unit 8 Quadratic Equations Homework 14. Each solution shows the reasoning, algebraic work, and final interpretation.

Problem 1 – Finding Maximum Height

A baseball is thrown upward from a height of 2 m with an initial velocity of 20 m/s. Write the height function, determine the time at which the ball reaches its maximum height, and calculate that maximum height.

Solution

1. Write the function
   [ h(t)= -\frac{1}{2}(9.8)t^{2}+20t+2 = -4.9t^{2}+20t+2. ]

2. Find the vertex time
   [ t_{max}= -\frac{b}{2a}= -\frac{20}{2(-4.9)} = \frac{20}{9.8}\approx 2.04\text{ s}. ]

3. Calculate maximum height
   [ h_{max}= -4.9(2.04)^{2}+20(2.04)+2. ]
   Compute stepwise:

   • ((2.04)^{2}=4.1616)
   • (-4.9\times4.1616 = -20.392)
   • (20\times2.04 = 40.8)
   • Sum: (-20.392+40.8+2 = 22.408) m.

   Answer: The ball reaches its peak at ≈ 2.04 s and the maximum height is ≈ 22.4 m.

Problem 2 – Time of Flight

A projectile is launched from the ground ((h_{0}=0)) with an initial upward velocity of 15 m/s. How long does it stay in the air before hitting the ground?

Solution

1. Height function
   [ h(t)= -4.9t^{2}+15t. ]

2. Set height to zero (launch and landing)
   [ -4.9t^{2}+15t = 0 ;\Longrightarrow; t(-4.9t+15)=0. ]

3. Solve for (t)

   • (t=0) (launch moment)
   • (-4.9t+15=0 ;\Rightarrow; t = \frac{15}{4.9}\approx 3.06) s.

   Answer: The projectile is airborne for ≈ 3.06 seconds.

Problem 3 – Finding Launch Height from Given Data

A ball reaches a maximum height of 30 m after 2.5 seconds. Assuming the only acceleration is gravity, find the initial height and

the initial velocity.

Solution

1. Use the vertex formula
   Since the vertex occurs at (t_{max}=2.5) s, we have
   [ t_{max} = -\frac{b}{2a} = 2.5. ]
   Here (a = -4.9) (gravity), so
   [ 2.5 = -\frac{b}{2(-4.9)} \quad\Rightarrow\quad b = 24.5. ]

2. Write the general height function
   [ h(t) = -4.9t^{2} + 24.5t + h_{0}. ]

3. Use the maximum height condition
   At (t=2.5), (h=30):
   [ 30 = -4.9(2.5)^{2} + 24.5(2.5) + h_{0}. ]
   Compute:

   • ((2.5)^{2}=6.25)
   • (-4.9\times6.25 = -30.625)
   • (24.5\times2.5 = 61.25)
   • Sum: (-30.625 + 61.25 + h_{0} = 30)
   • (30.625 + h_{0} = 30)
   • (h_{0} = -0.625) m.

4. Find initial velocity
   The coefficient (b) in the standard form (h(t) = -4.9t^{2} + v_{0}t + h_{0}) is the initial velocity (v_{0}). Thus (v_{0} = 24.5) m/s.

Answer: The ball was launched from a height of (-0.625) m (i.e., 0.625 m below the reference level, which could mean it was released from a point slightly below ground level in the model) with an initial upward velocity of 24.5 m/s.

Conclusion

Projectile motion is a classic application of quadratic equations, where the height of an object over time follows a parabola described by
[ h(t) = -\frac{1}{2}gt^{2} + v_{0}t + h_{0}. ]
By mastering the vertex formula, factoring, and the quadratic formula, you can quickly determine key features such as maximum height, time of flight, launch conditions, and landing time. These tools not only solve textbook problems but also model real‑world phenomena like sports, engineering trajectories, and physics experiments. With practice, interpreting the coefficients and extracting meaningful information from the quadratic model becomes second nature, turning abstract algebra into a powerful lens for understanding motion in the physical world.