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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3.4.11
Chapter 3, Problem 3.4.11

Finding g on a small airless planet Explorers on a small airless planet used a spring gun to launch a ball bearing vertically upward from the surface at a launch velocity of 15 m/sec. Because the acceleration of gravity at the planet’s surface was gₛ m/sec², the explorers expected the ball bearing to reach a height of s = 15t − (1/2)gₛt² m t sec later. The ball bearing reached its maximum height 20 sec after being launched. What was the value of gₛ?

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1
To find the value of gₛ, we need to determine when the ball bearing reaches its maximum height. The maximum height is reached when the velocity of the ball bearing is zero.
The velocity function v(t) is the derivative of the height function s(t) with respect to time t. So, differentiate s(t) = 15t - (1/2)gₛt² with respect to t to find v(t).
The derivative of s(t) with respect to t is v(t) = ds/dt = 15 - gₛt.
Set the velocity function v(t) to zero to find the time at which the maximum height is reached: 0 = 15 - gₛt.
Solve the equation 0 = 15 - gₛt for t = 20 seconds to find gₛ. This gives gₛ = 15/20.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration. In this problem, the equation s = 15t − (1/2)gₛt² represents the height of the ball bearing as a function of time, where 15 m/sec is the initial velocity and gₛ is the gravitational acceleration. Understanding these equations helps in determining the relationship between time, velocity, and acceleration.
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Maximum Height in Projectile Motion

The maximum height in projectile motion occurs when the vertical velocity of the object becomes zero. At this point, the object stops ascending and begins to descend. In the given problem, the ball bearing reaches its maximum height at 20 seconds, which is crucial for calculating the gravitational acceleration by setting the derivative of the height equation to zero.
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Solving Quadratic Equations

Solving quadratic equations is essential for finding unknown variables in kinematic problems. The height equation s = 15t − (1/2)gₛt² is quadratic in form. By setting the derivative of this equation to zero, we can solve for gₛ, the gravitational acceleration, using the known time at which the maximum height is reached.
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Related Practice
Textbook Question

Find all points on the curve y = tan x, −π/2 < x < π/2, where the tangent line is parallel to the line y = 2x. Sketch the curve and tangent lines together, labeling each with its equation.

Textbook Question

In Exercises 53 and 54, find dr/ds.


r cos 2s + sin²s = π

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For what value of c is the curve y = c/ (x + 1) tangent to the line through the points (0, 3) and (5, -2)?

Textbook Question

Derivatives in Differential Form


In Exercises 17–28, find dy.


y = cos(x²)

Textbook Question

Approximation Error


In Exercises 29–34, each function f(x) changes value when x changes from x₀ to x₀ + dx. Find


a. the change Δf = f(x₀ + dx) − f(x₀);

b. the value of the estimate df = fʹ(x₀) dx; and

c. the approximation error |Δf − df|.



f(x) = x⁴, x₀ = 1, dx = 0.1

Textbook Question

In Exercises 41–58, find dy/dt.


y = (1/6)(1 + cos²(7t))³