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We are given that the height \( s \) of the object as a function of time \( t \) is:

\[
s(t) = 192 - 16t^2
\]

where \( s \) is in feet, and \( t \) is in seconds. Let's break down and solve each part:

### a. Velocity, speed, and acceleration at time \( t \)

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

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 \( v(t) \) with respect to time:
   \[
   a(t) = \frac{dv}{dt} = \frac{d}{dt} (-32t) = -32 \text{ ft/s}^2
   \]
   This is a constant acceleration due to gravity (in the given units).

### b. Time to hit the ground

The object hits the ground when \( s(t) = 0 \). So, set the height equation equal to 0 and solve for \( t \):
\[
192 - 16t^2 = 0
\]
\[
16t^2 = 192
\]
\[
t^2 = 12
\]
\[
t = \sqrt{12} = 2\sqrt{3} \approx 3.464 \text{ seconds}
\]
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 \( v(t) = -32t \) and substituting \( t = 2\sqrt{3} \):
\[
v(2\sqrt{3}) = -32 \times 2\sqrt{3} = -64\sqrt{3} \approx -110.85 \text{ ft/s}
\]
The negative sign indicates the object is moving downward.

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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.

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?

Solve a kinematics problem where an object is dropped from a 192 ft tower. Learn how to calculate velocity, speed, and acceleration as a function of time, find the time to hit the ground, and determine the object's velocity at impact. This solution covers key physics concepts related to motion under gravity.

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