Final answer:
The velocity of the ball after 3 seconds of being dropped freely from the top of the Bunker Hill Monument is 96 feet per second. This is determined by taking the derivative of the given displacement function s(t) = 16t², which represents the distance in feet, to find the velocity formula v(t) = 32t, and then evaluating it at t = 3 seconds.
Step-by-step explanation:
To find the velocity of the ball after 3 seconds when dropped from the top of the Bunker Hill Monument, which is 221 feet tall, we use the given displacement equation s(t) = 16t² to calculate the displacement first and then determine the velocity.
The displacement function s(t) gives us the distance traveled by the ball after a certain time t seconds, when dropped freely from a certain height.
This specific form of the equation assumes acceleration due to gravity is approximately 32 feet per second squared (32 ft/s²).
As acceleration is the rate of change of velocity, the velocity function v(t) is the first derivative of the displacement function with respect to time.
Therefore, v(t) = s'(t) = d(16t²)/dt = 32t. To find the velocity after 3 seconds (t = 3s), inputting this time into the velocity function gives us: v(3) = 32× 3
= 96 ft/s.
So, the velocity of the ball after 3 seconds is 96 feet per second.