A ball is dropped from a 30-foot-tall building, meaning it has no initial velocity. Write a model h(t) that represents the height of the ball from the ground, in feet, t seconds after it is dropped from the building.(2 points) h(t)= t^2+ t+

Answer 1

To model the height of a ball dropped from a height of 30 feet with no initial velocity, we can use the following formula:

\[ h(t) = h_0 - \frac{1}{2} g t^2 \]

where:

• \( h(t) \) is the height of the ball at time \( t \),
• \( h_0 \) is the initial height (30 feet in this case),
• \( g \) is the acceleration due to gravity (approximately 32 feet per second squared on Earth),
• \( t \) is the time in seconds.

Since the ball is dropped from a height of 30 feet and has no initial upward velocity, we can substitute the values into the equation:

1. Initial height \( h_0 = 30 \) feet,
2. Acceleration due to gravity \( g \approx 32 \) ft/s².

The formula becomes:

\[ h(t) = 30 - \frac{1}{2} (32) t^2 \] \[ h(t) = 30 - 16t^2 \]

So the height of the ball as a function of time is:

\[ h(t) = 30 - 16t^2 \]

This means the height decreases as time increases until the ball reaches the ground (when \( h(t) = 0 \)).

Please let me know if you need further information!