Problem 71 - Algebra for College Students

Question

The height of a projectile fired vertically into the air (neglecting air resistance) at an initial velocity of 64 feet per second is a function of the time \((t)\) and is given by the equation \(h(t)=64 t-16 t^{2}\). Compute \(h(1), h(2), h(3)\), and \(h(4) . \quad 48 ; 64 ; 48 ; 0\)

Step-by-Step Solution

Answer

Heights are 48 ft, 64 ft, 48 ft, and 0 ft for \(t = 1, 2, 3, 4\) respectively.

Step 1: Substitute into the Equation

To find the height of the projectile at each second, substitute the value of \(t\) into the function \(h(t) = 64t - 16t^2\). Start with \(t = 1\).

Step 2: Calculate \(h(1)\)

Substitute \(t = 1\) into the equation:\[h(1) = 64(1) - 16(1)^2 = 64 - 16 = 48\]The height at \(t = 1\) second is 48 feet.

Step 3: Calculate \(h(2)\)

Next, substitute \(t = 2\) into the equation:\[h(2) = 64(2) - 16(2)^2 = 128 - 64 = 64\]The height at \(t = 2\) seconds is 64 feet.

Step 4: Calculate \(h(3)\)

Substitute \(t = 3\) into the equation:\[h(3) = 64(3) - 16(3)^2 = 192 - 144 = 48\]The height at \(t = 3\) seconds is 48 feet.

Step 5: Calculate \(h(4)\)

Finally, substitute \(t = 4\) into the equation:\[h(4) = 64(4) - 16(4)^2 = 256 - 256 = 0\]The height at \(t = 4\) seconds is 0 feet.

Key Concepts

Projectile Motion

When we talk about projectile motion, we're discussing an object that is projected into the air and is affected by gravity. The object follows a curved path, which is straightforward in the case of vertical motion without air resistance. The height of such an object over time can be described by a quadratic function, like the one given in the problem:

• The equation is in the form: \( h(t) = v_0 t - \, 16 t^2 \).
• The term \( v_0 t \) represents the initial velocity multiplied by time.
• The \( 16 t^2 \) term accounts for the downward acceleration due to gravity, approximated as 32 feet per second squared, halved because height is the focus, not distance.

Understanding projectile motion helps in many real-world applications, such as sports and engineering.

Velocity

Velocity is key when discussing projectile motion. It tells us how fast an object is moving and in what direction. In this exercise, the initial velocity is 64 feet per second upwards.

• This initial velocity is what propels the projectile into the air.
• Negative velocity results from the gravitational pull, which eventually brings the projectile back down to the ground.

When the problem posed the equation \( h(t) = 64t - 16t^2 \), the 64 represents the initial velocity, showing its effect on each second of travel. The concept of velocity not only defines speed but also plays a significant role in the maximum height the projectile reaches.

Time

Time in this context measures how long the projectile maintains its motion. Calculating the height at different times gives insight into how the projectile moves through air. This exercise directly relates to the concept as follows:

• Heights at specific times (\( t = 1, 2, 3, 4 \)) are given to show the progression of the projectile's motion.
• As time progresses, gravity affects the projectile more, decreasing its height after reaching the peak.

Understanding the role of time helps when predicting where and when the projectile will land. Given the equation, substituting different time intervals provides a clear view of the projectile's trajectory.

Substitution Method

This is a standard method in solving mathematics problems where you replace variables with specific values. In this exercise, we use substitution to find the projectile's height at different times:

• Substituting \( t = 1 \), \( t = 2 \), \( t = 3 \), and \( t = 4 \) helps us calculate the corresponding heights.
• By substituting different time values, the equation \( h(t) = 64t - 16t^2 \) becomes specific to that instant.

Using substitution is particularly efficient here as it simplifies the calculation of a quadratic function for discrete time intervals, giving instantaneous height readings. This method ensures accurate, straightforward results for each timing checkpoint in the motion.