Question

A ball is thrown directly upward from a height of 8 ft with an initial velocity of 20 ​ft/sec. The function ​s(t) = −16t²+20t+8 gives the height of the​ ball, in​ feet, t seconds after it has been thrown. Determine the time at which the ball reaches its maximum height and find the maximum height.

Answer 1

Answer:

Step-by-step explanation:

I suppose  we could factor this and find out how long the ball was in the air. From there we could determine its halfway point in terms of time, and then sub that time in for t in the position function to get the height at that time. But in order to avoid that, which may actually lead to an estimation as opposed to the actual height and time, we will use calculus.

Keep in mind that the first derivative of the position function is velocity. You have learned in Physics that if an object is at the very tip-top of its travels it has 0 velocity (this is because the object HAS to stop moving in order to turn around and head the other direction). So we will simply find the function's derivative, set it equal to 0 and then solve for t. The position function is

$s(t)=-16t^2+20t+8$

The first derivative, aka as the velocity function, is

$v(t)=-32t+20$

Setting it equal to 0:

-20 = -32t so

$t=\frac{5}{8}sec$

That is WHEN the object is at its max height. To find out what the max height it, we will sub that t value into the position function.

$s(\frac{5}{8})=-16(\frac{5}{8})^2+20(\frac{5}{8})+8$ gives us that the ball's position at five-eighths of a second is 14.25 feet.