Problem 15 - Algebra A Combined Function

Question

A projectile is fired upward from the ground with an initial velocity of 300 feet per second. Neglecting air resistance, the height of the projectile at any time \(t\) can be described by the polynomial function \(P(t)=-16 t^{2}+300 t\). Find the height of the projectile at each given time. a. \(t=1\) second b. \(t=2\) seconds c. \(t=3\) seconds d. \(t=4\) seconds e. Explain why the height increases and then decreases as time passes. f. Approximate (to the nearest second) how long before the object hits the ground.

Step-by-Step Solution

Answer

a. 284 feet, b. 536 feet, c. 756 feet, d. 944 feet; e. Gravity slows it down. f. Hits the ground in 19 seconds.

Step 1: Height calculation for t=1 second

Insert the value of \( t = 1 \) into the polynomial function \( P(t) = -16t^2 + 300t \). Compute as follows: \[ P(1) = -16(1)^2 + 300(1) = -16 + 300 = 284 \] Thus, the height of the projectile at \( t = 1 \) second is 284 feet.

Step 2: Height calculation for t=2 seconds

Insert the value of \( t = 2 \) into the polynomial function. Compute as follows: \[ P(2) = -16(2)^2 + 300(2) = -64 + 600 = 536 \] Thus, the height of the projectile at \( t = 2 \) seconds is 536 feet.

Step 3: Height calculation for t=3 seconds

Insert the value of \( t = 3 \) into the polynomial function. Compute as follows: \[ P(3) = -16(3)^2 + 300(3) = -144 + 900 = 756 \] Thus, the height of the projectile at \( t = 3 \) seconds is 756 feet.

Step 4: Height calculation for t=4 seconds

Insert the value of \( t = 4 \) into the polynomial function. Compute as follows: \[ P(4) = -16(4)^2 + 300(4) = -256 + 1200 = 944 \] Thus, the height of the projectile at \( t = 4 \) seconds is 944 feet.

Step 5: Explanation of height variation

The height of the projectile increases initially due to the upward velocity. As time increases, gravity begins to dominate, causing a decrease in height. This is because the term \(-16t^2\) represents the downward acceleration due to gravity, which increases as time passes.

Step 6: Time for projectile to hit the ground

To find when the projectile hits the ground, set \( P(t) = 0 \) and solve for \( t \):\[ -16t^2 + 300t = 0 \]Factor the equation:\[ t(-16t + 300) = 0 \]This gives solutions \( t = 0 \) (when the projectile is launched) and \(-16t + 300 = 0 \).Solve for \( t \):\[ -16t = -300 \]\[ t = \frac{300}{16} = 18.75 \]Therefore, the projectile will hit the ground approximately 19 seconds after being launched.

Key Concepts

Polynomial Functions

In projectile motion, the height of the projectile over time is often described using polynomial functions. A polynomial function encompasses expressions of multiple terms, relying on powers of a variable like time. Here, the function is given by \(P(t) = -16t^2 + 300t\). This specific function is of degree two because the highest power of the variable \(t\) is 2.

• The term \(-16t^2\) represents the effects of gravity acting downwards on the projectile.
• The term \(300t\) represents the initial velocity propelling the projectile upwards.

The behavior of polynomial functions, particularly quadratics, shows us how various factors interact, such as initial velocity, directions, and more. Quadratic polynomials like this one help express how projectile motion is not just a mere upward or downward journey, but a smooth arc eventually.
When you plug different times into this function, it provides the height of the projectile at those specific moments.

Gravity Effects

One of the crucial concepts in projectile motion is the effect of gravity. The negative sign of the term \(-16t^2\) in the polynomial \(P(t) = -16t^2 + 300t\) indicates gravity's downward pull. Gravity constantly accelerates the projectile downwards at a rate of \(32 \, \text{ft/s}^2\), responsible for this specific coefficient.

• As the projectile rises, gravity slows it down, gradually decreasing its upward speed.
• Once the projectile reaches its peak height, gravity then accelerates it downward, increasing its speed in the descent.

This creates a parabolic trajectory that is typical of projectile motion. The impact of gravity is often simplified in problems to make calculations more straightforward, assuming air resistance is negligible.
Understanding gravity's role is critical in effectively predicting the motion path and timing of the projectile.

Quadratic Equations

Quadratic equations are key when dealing with projectile problems involving polynomial functions of degree two, like our example function \(P(t) = -16t^2 + 300t\). Solving a quadratic equation involves finding the values of \(t\) where the expression equals zero, which tells us important information about projectile motion, such as when an object hits the ground.
To solve \(-16t^2 + 300t = 0\):

• Factor the equation: \(t(-16t + 300) = 0\).
• This reveals two potential solutions: \(t = 0\) (when the projectile is launched) and another solution where \(-16t + 300 = 0\).
• Solve for \(t\) to find when the projectile will return to the ground: \(t = \frac{300}{16} = 18.75\).

Thus, we establish that the projectile will likely hit the ground approximately 19 seconds after its launch. Quadratic equations capture the symmetry and transformative aspects of parabolic paths, providing profound insights into the timing and trajectory in projectile motion.