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.
h(t)=_t^2+_t+_
h(t) = -16t^2 + 30
To write the model h(t) that represents the height of the ball from the ground, we can use the equation for motion under constant acceleration.
Since the ball is dropped from rest, its initial velocity (u) is zero. The acceleration due to gravity (g) is approximately 32 feet per second squared.
The equation for the height of the ball from the ground at time t seconds after it is dropped can be written as:
h(t) = ut + (1/2)gt^2
Now, substituting the values for u and g into the equation, we have:
h(t) = (1/2)gt^2
Therefore, the model representing the height of the ball from the ground after t seconds is:
h(t) = (1/2)gt^2
In this case, the acceleration due to gravity (g) is approximately 32 feet per second squared. Thus, the final model is:
h(t) = (1/2)(32)t^2
Simplifying further, we get:
h(t) = 16t^2
So, the model representing the height of the ball from the ground (in feet) after t seconds is h(t) = 16t^2.