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If a ball is thrown into the air with a velocity of $ 40 ft/s $, its height (in feet) after $ t $ seconds is given by $ y = 40t - 16t^2 $. Find the velocity when $ t = 2 $.

$-24 f t / s e c$

04:12

Daniel J.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 7

Derivatives and Rates of Change

Limits

Derivatives

Missouri State University

University of Michigan - Ann Arbor

University of Nottingham

Idaho State University

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So here we give an example of a certain function where we have Y equals 40 t minus 16 t squared. And were given information that this is our position function, And the derivative of the position function gets the velocity function, which is 40 -32 t. So this is our velocity function And were asked to evaluate the velocity at time of T equals two seconds. So why prime of two would be equivalent to 40 64? So this would be equivalent to negative 24? And our units is specifically in this case feet per second, and this basically shows that one and the ball is thrown into the air at time T equals two seconds. This is giving the part of basically the falling face of the ball, so it's after the ball has already reached its peak, and this is our final answer.

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