Problem 76 - College Algebra

Question

Use the position function $$ s(t)--16 t^{2}+v_{0} t+s_{0} $$ \(\left(v_{11}=\text { initial velocity, } s_{0}-\text { initial position, } t-\text { time }\right)\) to answer Exercises \(75-76\) You throw a ball straight up from a rooftop 160 feet high with an initial velocity of 48 feet per second. During which time period will the ball's height exceed that of the rooftop?

Step-by-Step Solution

Answer

The ball's height exceeds that of the rooftop from just after time \(t=0\) until just before time \(t=3\) seconds.

Step 1: Understand and identify given quantities

The initial height \(s_0\) is 160 feet. The initial upward velocity \(v_0\) is 48 feet per second. The acceleration \(a\) is -16 feet/second squared (due to gravity). We aim to find the time \(t\) when the ball's height is greater than the rooftop.

Step 2: Write the height equation

The height \(s(t)\) of the ball at any time \(t\) can be described by the equation: \(s(t)=-16t^2+v_0t+s_0\). Substituting the given values, our equation becomes: \(s(t) = -16t^2 + 48t + 160\).

Step 3: Solve the equation

We need to solve the equation \(s(t) = 160\). Setting \(s(t)\) equal to 160 and simplifying, we get: \(-16t^2 + 48t + 160 = 160\). This simplifies to \(-16t^2 + 48t = 0\). Solve this equation for \(t\). The solutions are \(t=0\) and \(t=3\) seconds. This means the ball's height is equal to the rooftop at \(t=0\) (the start) and \(t=3\) seconds (when it comes back down).

Step 4: Determine the time period

So, the ball's height exceeds the rooftop's height for \(t\) values in between \(0\) and \(3\) seconds (not including these endpoints). Therefore, the ball's height exceeds that of the rooftop from just after time \(t=0\) until just before time \(t=3\) seconds.

Key Concepts

Position Function

In physics, the position function describes the location of an object at a given time. When you throw a ball into the air, its height can be calculated using a specific equation. For our exercise, the position function is given by:
\[s(t) = -16t^2 + v_0 t + s_0\]Here,

• \(s(t)\) is the height at time \(t\).
• \(v_0\) is the initial velocity.
• \(s_0\) is the initial position.
• The term \(-16t^2\) represents the effect of gravity on the position.

Understanding this function helps in analyzing how the initial velocity and gravity affect the height of an object.

Initial Velocity

Initial velocity refers to the speed at which an object starts its motion. In our case, the ball has an initial velocity of 48 feet per second. This means at the moment you release it, the ball is moving upwards at this speed. Initial velocity is crucial because:

• It determines how high the ball will travel before gravity slows it down.
• The greater the initial velocity, the higher and longer the object will remain above its starting point.

In our exercise, the positive initial velocity indicates upward motion against the pull of gravity.

Gravity

Gravity is the force that pulls objects towards the Earth. In the position function, it's represented by the term \(-16t^2\). Here's why gravity is significant:

• It reduces the upward velocity of the ball, causing it to slow down, stop, and then fall back.
• The "-16" in the equation means gravity causes the object to accelerate downwards at 16 feet per second squared.

Understanding gravity's role can help predict when the ball will stop rising and start falling. This is why the ball eventually returns to the rooftop height at a later time in the motion.

Time Period Analysis

Analyzing the time period involves determining when the ball's height is greater than its starting point. In our problem, you need to find when the height \(s(t)\) exceeds 160 feet (the rooftop). By solving the equation \[-16t^2 + 48t + 160 = 160\]You find that the ball goes higher than the rooftop for times between \(t = 0\) and \(t = 3\) seconds. Let's break it down:

• The ball starts at the rooftop height at \(t = 0\).
• It ascends, reaching higher than 160 feet soon after \(t = 0\).
• It exceeds this height until just before it returns to rooftop level at \(t = 3\).

So, the ball exceeds the rooftop height from just after launch until just before it lands back.