Math Problem Statement

An object is dropped from a tower, 192 ft above the ground. The object's height above ground t sec into the fall is s = 192 − 16t^2. a. What is the object's velocity, speed, and acceleration at time t? b. About how long does it take the object to hit the ground? c. What is the object's velocity at the moment of impact?

Solution

We are given that the height ss of the object as a function of time tt is:

s(t)=19216t2s(t) = 192 - 16t^2

where ss is in feet, and tt is in seconds. Let's break down and solve each part:

a. Velocity, speed, and acceleration at time tt

  1. Velocity is the derivative of the position function s(t)s(t) with respect to time: v(t)=dsdt=ddt(19216t2)=32tv(t) = \frac{ds}{dt} = \frac{d}{dt} \left( 192 - 16t^2 \right) = -32t

  2. Speed is the magnitude of velocity: \text{Speed} = |v(t)| = |-32t| = 32t \quad \text{(since time is positive, speed is 32t)}.

  3. Acceleration is the derivative of the velocity function v(t)v(t) with respect to time: a(t)=dvdt=ddt(32t)=32 ft/s2a(t) = \frac{dv}{dt} = \frac{d}{dt} (-32t) = -32 \text{ ft/s}^2 This is a constant acceleration due to gravity (in the given units).

b. Time to hit the ground

The object hits the ground when s(t)=0s(t) = 0. So, set the height equation equal to 0 and solve for tt: 19216t2=0192 - 16t^2 = 0 16t2=19216t^2 = 192 t2=12t^2 = 12 t=12=233.464 secondst = \sqrt{12} = 2\sqrt{3} \approx 3.464 \text{ seconds} Thus, it takes approximately 3.46 seconds for the object to hit the ground.

c. Velocity at the moment of impact

Using the velocity equation v(t)=32tv(t) = -32t and substituting t=23t = 2\sqrt{3}: v(23)=32×23=643110.85 ft/sv(2\sqrt{3}) = -32 \times 2\sqrt{3} = -64\sqrt{3} \approx -110.85 \text{ ft/s} The negative sign indicates the object is moving downward.


Do you want further details or have any questions?

Here are 5 related questions to expand understanding:

  1. How does air resistance affect the equations of motion for falling objects?
  2. What would the velocity and time to hit the ground be if the object were thrown downward with an initial velocity?
  3. How can we determine the maximum height of a projectile if it is launched upward instead?
  4. What happens to the time and velocity if the object falls from a different height?
  5. How does the acceleration due to gravity change on different planets?

Tip: The speed of a falling object increases linearly with time under constant acceleration, while the distance covered increases quadratically.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Differentiation
Quadratic Equations

Formulas

s(t) = 192 - 16t^2
v(t) = -32t (Velocity)
a(t) = -32 (Acceleration)

Theorems

Constant Acceleration due to Gravity
Kinematic Equations

Suitable Grade Level

Grades 10-12