Math Problem Statement
Solution
We are given that the height of the object as a function of time is:
where is in feet, and is in seconds. Let's break down and solve each part:
a. Velocity, speed, and acceleration at time
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Velocity is the derivative of the position function with respect to time:
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Speed is the magnitude of velocity: \text{Speed} = |v(t)| = |-32t| = 32t \quad \text{(since time is positive, speed is 32t)}.
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Acceleration is the derivative of the velocity function with respect to time: This is a constant acceleration due to gravity (in the given units).
b. Time to hit the ground
The object hits the ground when . So, set the height equation equal to 0 and solve for : 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 and substituting : 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:
- How does air resistance affect the equations of motion for falling objects?
- What would the velocity and time to hit the ground be if the object were thrown downward with an initial velocity?
- How can we determine the maximum height of a projectile if it is launched upward instead?
- What happens to the time and velocity if the object falls from a different height?
- 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
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