The height above ground of an object moving vertically is...

The height above ground of an object moving vertically is given by $$s=-16 t^{2}+96 t+112$$ with $s$ in feet and $t$ in seconds. Find a. the object's velocity when $t=0$; b. its maximum height and when it occurs; c. its velocity when $s=0$.

Transcript

00:01 Okay, we are going to start with a position function, s, which is given in feet, equal to negative 16 t squared plus 96t plus 112, and t is time in seconds.

00:29 And so the first thing we want to do is to determine the velocity at t equal to zero.

00:59 And so velocity of a position function is given by the first derivative.

01:10 So this will be negative 32t plus 96.

01:14 And so i want to evaluate that velocity function at time zero, which gives me negative, 96, not negative, 96 feet per second.

01:30 Now what we want to do, part b is find the max height and the time associated with that maximum height.

01:46 Well, maximum or minimum occurs at the derivative, at the critical values of the derivative.

01:56 And so we're going to take that derivative, which is negative 32t plus 96, and set it equal to zero, and solve for that time.

02:11 And so we get a time value of three seconds.

02:18 Okay.

02:19 So now does that time is that it's a critical value.

02:25 So we know that a relative max or relative men, sometimes known as a local max or a local man occurs at that time.

02:33 But is it a maximum or would it be associated with a max height or a men height? so we're going to take that second derivative, which is negative 32, which tells me that for all time, the graph is concave down.

03:02 So t equal to three seconds is a place of a local, uh, local...

Question

The height above ground of an object moving vertically is given by $$s=-16 t^{2}+96 t+112$$ with $s$ in feet and $t$ in seconds. Find a. the object's velocity when $t=0$; b. its maximum height and when it occurs; c. its velocity when $s=0$.

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