Problem 141 - Calculus Volume 1

Question

A ball is thrown into the air and its position is given by \(h(t)=-4.9 t^{2}+60 t+5 \mathrm{m} .\) Find the height at which the ball stops ascending. How long after it is thrown does this happen?

Step-by-Step Solution

Answer

The ball stops ascending 6.12 seconds after it's thrown, at a height of approximately 188.4 meters.

Step 1: Understanding the Problem

We need to find the point at which the ball stops ascending. This is the point when its velocity becomes zero. The height of the ball as a function of time is given as \(h(t)=-4.9t^2+60t+5\).

Step 2: Calculating the Velocity

To find when the ball stops ascending, we need to calculate the derivative of the height function to find the velocity. The velocity \(v(t)\) is given by \(v(t) = \frac{dh}{dt} = -9.8t + 60\).

Step 3: Finding When the Ball Stops Ascending

Set the velocity equation equal to zero and solve for \(t\), i.e., solve \(-9.8t + 60 = 0\). This gives \(t = \frac{60}{9.8}\).

Step 4: Calculating the Precise Time

Perform the division to find the exact value of \(t\). \(t = \frac{60}{9.8} \approx 6.12\) seconds. This is the time when the ball stops ascending.

Step 5: Evaluating the Height at Maximum

Substitute \(t = 6.12\) back into the height equation to find the height at this time: \(h(6.12) = -4.9(6.12)^2 + 60(6.12) + 5\).

Step 6: Conclusion

Calculate \(h(6.12)\) to get the maximum height, which is approximately 188.4 meters.

Key Concepts

Derivative of a Function

The derivative of a function indicates how the function's output changes concerning a change in input. In calculus, derivatives are essential for understanding the behavior of functions.
For motion problems, like the one involving the ball thrown in the air, the derivative helps us find the velocity from the position function.
Given the height function of the ball, represented as \(h(t) = -4.9t^2 + 60t + 5\), the derivative with respect to time \(t\), gives us the velocity function. This means we're looking at how the height changes over time, which is:

• If the derivative (velocity) is positive, the object is moving upwards.
• If the derivative (velocity) is zero, the object is momentarily stationary (at the peak height).
• If the derivative (velocity) is negative, the object is moving downwards.

By taking the derivative of the height function, we get:\[v(t) = \frac{dh}{dt} = -9.8t + 60\].
This equation represents the ball's velocity at any time \(t\). Understanding how to calculate derivatives allows us to predict motion dynamics accurately and solve real-world problems like finding the point where a ball stops ascending.

Velocity Calculation

Velocity shows how fast an object changes its position over time. For the ball thrown into the air, velocity calculation comes from the derivative of its height function.
The function for height is given by: \(h(t) = -4.9t^2 + 60t + 5\).
By differentiating, we find the velocity function: \(v(t) = -9.8t + 60\).
This indicates how fast the ball moves at any given time moment \(t\).To find when the ball stops moving upwards, you're solving \(v(t) = 0\).
This situation occurs when:\(-9.8t + 60 = 0\).

• Solving for \(t\) gives \(t = \frac{60}{9.8} \approx 6.12\) seconds.

This means the ball stops ascending approximately 6.12 seconds after it has been thrown. Understanding velocity is crucial because it also helps in determining the moment an object changes its direction which is important for predicting future positions and movements.

Maximum Height

Maximum height is the peak point an object reaches in its vertical motion. At this point, the velocity becomes zero as the object stops ascending briefly before descending.
In this exercise, we first determined the time at which the ball reaches its maximum height, which was around 6.12 seconds (calculated from velocity being zero).
Next, to find the maximum height, we substitute \(t = 6.12\) back into the initial height function:
\[h(6.12) = -4.9(6.12)^2 + 60(6.12) + 5\].

• Calculate the value to find the maximum height, which is approximately 188.4 meters.

Knowing the maximum height is essential for many practical reasons such as ensuring safety margins for flyover passages or predicting landing spots in projectile motion.
Through calculus, determining maximum height teaches us about the natural behavior of objects in motion and helps us make necessary adjustments or predictions in real-world applications.