Question

if a ball is thrown into the air with a velocity of 39 ft/s, its height (in feet) after t seconds is given by y = 39t - 16t2. find the velocity when t = 2.

Answer 1

To find the velocity of the ball at two seconds, differentiate the position function and then substitute t = 2 into the velocity function, yielding -25 ft/s, indicating a downward motion.

Step-by-step explanation:

To find the velocity of a ball at a specific time when it is thrown into the air, we can use calculus to differentiate the equation y = 39t - 16t2, which gives us the velocity function. The derivative of the position function y with respect to time t gives us the velocity function v(t). Differentiating 39t with respect to t gives 39, and differentiating -16t2 gives -32t. Therefore, the velocity function is v(t) = 39 - 32t. To find the velocity when t = 2, substitute 2 for t into the velocity function, which results in v(2) = 39 - 32(2) = 39 - 64 = -25 ft/s. The negative sign indicates that the ball is moving downwards at 2 seconds after being thrown.