Problem 62

Question

A toy rocket is launched from the top of a building 50 feet tall at an initial velocity of 200 feet per second. Let $t$ represent the amount of time elapsed after the launch. (a) Express the height $s$ as a function of the time $t$ (b) Determine both analytically and graphically the time at which the rocket reaches its highest point. How high will it be at that time? (c) For what time interval will the rocket be more than 300 feet above ground level? Determine the answer graphically, and give times to the nearest tenth of a second. (d) After how many seconds will the rocket hit the ground? Determine the answer graphically.

Step-by-Step Solution

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Answer

(a) Height function: $ s(t) = -16t^2 + 200t + 50 $. (b) Highest point at 6.25 seconds, height: 675 feet. (c) Above 300 feet between 1.4 and 11.4 seconds. (d) Hits ground at 12.7 seconds.

1Step 1: Define the equation for height as a function of time

The height of the toy rocket as a function of time can be found using the equation for vertical motion. The standard form is $ s(t) = -16t^2 + v_0t + h_0 $, where $ v_0 $ is the initial velocity, and $ h_0 $ is the initial height. For this problem, $ v_0 = 200 \text{ ft/s} $ and $ h_0 = 50 \text{ ft} $. Thus, the equation becomes $ s(t) = -16t^2 + 200t + 50 $. This will give the height at any time $ t $.

2Step 2: Find the time at which the rocket reaches its highest point analytically

To find the time when the rocket is at its highest point, determine the vertex of the parabola represented by the equation $ s(t) = -16t^2 + 200t + 50 $. The time $ t $ at the vertex can be calculated using the formula $ t = -\frac{b}{2a} $, where $ a = -16 $ and $ b = 200 $. Substituting these values, $ t = -\frac{200}{2(-16)} = 6.25 \text{ seconds} $. Thus, the rocket reaches the highest point at 6.25 seconds.

3Step 3: Calculate the maximum height of the rocket

To determine the maximum height, substitute $ t = 6.25 $ into the height function $ s(t) = -16t^2 + 200t + 50 $. This results in $ s(6.25) = -16(6.25)^2 + 200(6.25) + 50 $. Simplifying gives $ s(6.25) = 675 \text{ feet} $. Thus, the maximum height of the rocket is 675 feet.

4Step 4: Determine the time interval when the rocket is more than 300 feet above ground

To find when the rocket is more than 300 feet above the ground, set the equation $ s(t) = -16t^2 + 200t + 50 $ greater than 300. Solving $ -16t^2 + 200t + 50 > 300 $ leads to $ -16t^2 + 200t - 250 > 0 $. This quadratic equation can be solved for the roots, which are $ t_1 \approx 1.395 $ and $ t_2 \approx 11.36 $. Thus, the rocket is above 300 feet between $ t = 1.4 $ seconds and $ t = 11.4 $ seconds (rounded to the nearest tenth).

5Step 5: Find when the rocket hits the ground

The rocket hits the ground when its height $ s(t) = 0 $. Solving the equation $ -16t^2 + 200t + 50 = 0 $, we find the time it takes to hit the ground. Using the quadratic formula gives $ t \approx -0.24 \text{ seconds} $ (non-physical solution) and $ t \approx 12.74 \text{ seconds} $. Therefore, the rocket hits the ground at about 12.7 seconds.

Key Concepts

Quadratic FunctionsMaximum HeightTime of FlightInequalities

Quadratic Functions

Quadratic functions are mathematical expressions typically seen in the form of $ ax^2 + bx + c $. They graph into a U-shaped curve called a parabola. In projectile motion problems, like the toy rocket problem here, quadratic functions are invaluable.

When dealing with vertical motion, the height $ s(t) $ of an object at time $ t $ is often modeled by a quadratic function. For example, the function $ s(t) = -16t^2 + 200t + 50 $ describes the rocket’s height over time, where:

$-16t^2$ represents the effect of gravity (in feet per second squared).
$200t$ is the initial velocity term (feet per second).
$50$ is the initial height from where the rocket was launched in feet.

This function allows us to predict the height at any given time during the flight.

Maximum Height

To find the maximum height reached by the rocket, we need to identify the vertex of the parabola formed by the quadratic function $ s(t) = -16t^2 + 200t + 50 $. The vertex formula $ t = -\frac{b}{2a} $ helps us find the point of maximum or minimum of a parabola.

With $ a = -16 $ and $ b = 200 $, we have:

$ t = -\frac{200}{2 \times (-16)} = 6.25 $ seconds.

By substituting $ t = 6.25 $ back into the original equation, we calculate:

$ s(6.25) = -16(6.25)^2 + 200(6.25) + 50 $.
This equals a maximum height of $ 675 $ feet.

Understanding how to calculate the vertex of a parabola helps us determine such crucial points in projectile motion.

Time of Flight

The time of flight in a projectile motion refers to how long an object is in the air. For the toy rocket, we determine this by finding when the rocket hits the ground, i.e., when the height $ s(t) $ is zero.

Using the quadratic formula on the equation $ -16t^2 + 200t + 50 = 0 $, the times $ t $ are

$ t \approx -0.24 $ seconds (which isn't physically meaningful since time cannot be negative).
$ t \approx 12.74 $ seconds, meaning the rocket's total flight duration before returning to the ground is approximately $ 12.7 $ seconds.

The time of flight is essential in determining the complete duration in which the object is moving through the air.

Inequalities

Inequalities are extremely useful when assessing specific conditions of projectile motion, such as determining when the rocket is above a certain height. Here, we use the inequality $ -16t^2 + 200t + 50 > 300 $ to find when the rocket is more than 300 feet above ground.

Simplifying to $ -16t^2 + 200t - 250 > 0 $, we solve the quadratic inequality and determine the roots:

The approximate roots are $ t_1 \approx 1.395 $ and $ t_2 \approx 11.36 $.

This means the rocket stays above 300 feet between $ 1.4 $ seconds and $ 11.4 $ seconds.

Inequalities like this one give precise insights into the time intervals where certain conditions are met.

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