Problem 39 - Brief Applied Calculus

Question

Maximum Height If a bullet from a 9 -millimeter pistol is fired straight up from the ground, its height \(t\) seconds after it is fired will be \(s(t)=-16 t^{2}+1280 t\) feet (neglecting air resistance) for \(0 \leq t \leq 80\). a. Find the velocity function. b. Find the time \(t\) when the bullet will be at its maximum height. [Hint: At its maximum height the bullet is moving neither up nor down, and has velocity zero. Therefore, find the time when the velocity \(v(t)\) equals zero.] c. Find the maximum height the bullet will reach. [Hint: Use the time found in part (b) together with the height function \(s(t) .]\)

Step-by-Step Solution

Answer

The bullet reaches its maximum height of 25600 feet at 40 seconds.

Step 1: Find the Velocity Function

The velocity of an object is the derivative of its position function with respect to time. Since we have the height function as \(s(t) = -16t^2 + 1280t\), the velocity function \(v(t)\) can be found by differentiating \(s(t)\) with respect to \(t\). The derivative of \(-16t^2\) is \(-32t\) and the derivative of \(1280t\) is \(1280\). So, \(v(t) = -32t + 1280\).

Step 2: Determine the Time of Maximum Height

To find the time when the bullet reaches its maximum height, we set the velocity function equal to zero since the velocity is zero at the peak of its trajectory. So, we solve \(-32t + 1280 = 0\). Add \(32t\) to both sides to get \(1280 = 32t\). Divide both sides by \(32\) to find \(t = 40\) seconds.

Step 3: Calculate the Maximum Height

Now that we know the bullet reaches its maximum height at \(t = 40\) seconds, we substitute \(t = 40\) into the height function \(s(t) = -16t^2 + 1280t\) to find the maximum height. Substituting gives \(s(40) = -16(40)^2 + 1280(40)\). First compute \(40^2 = 1600\), then \(-16 \times 1600 = -25600\) and \(1280 \times 40 = 51200\). So, \(s(40) = -25600 + 51200 = 25600\) feet.

Key Concepts

Derivative

The derivative is a crucial concept in calculus, especially when it comes to understanding motion. The derivative of a function gives us the rate at which that function is changing. When dealing with the height of an object over time, the derivative of this height function will give us its velocity.

For example, in the given problem, the height function is given as \( s(t) = -16t^2 + 1280t \). To find the velocity function, which tells us how quickly the height is changing, we need to calculate the derivative of \( s(t) \) with respect to \( t \).

• The derivative of \(-16t^2\) is \(-32t\).
• The derivative of \(1280t\) is \(1280\).

Thus, the velocity function \( v(t) \) is \(-32t + 1280\). This function tells us the speed and direction of the bullet at any time \( t \). The negative coefficient of \( t \) in the velocity function shows that the velocity decreases as time increases, which is exactly what we would expect for an object moving upwards against gravity.

Velocity Function

The velocity function plays an important role in determining how the height of the bullet changes over time. It tells us whether the bullet is moving up or down and how fast it is moving at any given time.

In our example, the velocity function is \( v(t) = -32t + 1280 \). This linear function decreases as time increases due to the term \(-32t\). As time progresses, the upward velocity decreases until it reaches zero at the bullet's maximum height. This is because gravity continuously decelerates the bullet until it momentarily stops before beginning its downward descent.

• When \( v(t) > 0 \), the bullet is moving upwards.
• When \( v(t) = 0 \), the bullet is at its peak.
• When \( v(t) < 0 \), the bullet is descending.

By setting the velocity function equal to zero \( (-32t + 1280 = 0) \), we solve for \( t \) and find that the bullet reaches its highest point at \( t = 40 \) seconds.

Maximum Height

The maximum height of an object in motion is a point of interest in many real-world applications, as it represents the peak of the object's ascent. In the problem given, once we establish the time when the velocity is zero using the velocity function, we can easily find the maximum height using the original height function.

After determining from the velocity function that the maximum height occurs at \( t = 40 \) seconds, we substitute \( t = 40 \) into the height function \( s(t) = -16t^2 + 1280t \) to find the bullet's maximum height.

• Calculate \( 40^2 \) which is \(1600\).
• Compute \(-16 imes 1600 = -25600 \).
• Determine \(1280 imes 40 = 51200 \).

Adding these results, we find the maximum height of the bullet is \( 25600 \) feet.

Understanding how to find the maximum height using calculus is vital, especially in fields like engineering and physics, as it allows for the precise modeling and prediction of an object's behavior under various forces.