If a ball is thrown vertically upward with a velocity of...

Key Concepts

Quadratic Functions

When an object is projected vertically under constant acceleration, its displacement as a function of time is modeled by a quadratic equation. The parabolic shape of this function helps in identifying critical features such as the maximum height, which corresponds to the vertex of the parabola. Understanding the properties of quadratic functions, including vertex calculation, is essential for analyzing such motions.

Derivatives and Kinematics

In kinematics, the derivative of the displacement function with respect to time gives the velocity function. This connection allows one to determine the instantaneous velocity of an object at any point along its path. Using differentiation, one can assess key instances of motion such as the moment when the object's velocity is zero at maximum height or when it takes specific values at given displacements.

Projectile Motion

Projectile motion is the motion of an object thrown or projected into the air, subject only to the acceleration of gravity. It deals with the analysis of both the horizontal and vertical components of motion. In vertical projectile motion, the object experiences a constant downward acceleration due to gravity, which is a key factor in predicting its height and velocity at any given time.

Uniform Acceleration

Uniform acceleration implies that the rate of change of velocity is constant. In the context of projectile motion, this constant acceleration is typically due to gravity, and it simplifies the analysis of motion through the use of kinematic equations. These equations relate displacement, initial velocity, time, and acceleration, allowing for the prediction of quantities such as maximum height and velocity at specific displacements.

Transcript

00:01 Okay, so for this problem, we're given that if we throw a ball vertically upward, then this is going to be our position function.

00:09 So we're asked at what is the maximum height reached by the ball, and then we're asked, what is the velocity of the ball when the height is 29 .4 on its way up and on its way down.

00:25 And so let's start with part a first.

00:28 So we're asked about the maximum height.

00:33 And so whenever i hear the word maximum, i'm automatically thinking first derivative, because that gives us our maximum point.

00:41 And in this case, it wouldn't be the maximum height, but it would give us the time, t, at which the ball it has a maximum.

00:53 And so then we'll take that t and plug it into our position function to get the height.

00:59 So first up, we're going to do.

01:01 To do s prime of t, which is going to be 24 .5 minus 9 .8t.

01:09 And when we're wanting to find maximum or minimum, we want to find critical points.

01:14 So we're going to set this equal to zero.

01:17 So then we get 24 .5 is equal to 9 .8t.

01:22 Then we get t is going to be equal to 2 .5.

01:28 So now that we have found our t value.

01:35 We're not quite done because this is where our t is the maximum.

01:39 And actually, we do in fact need to check that this is the maximum.

01:44 Let's just do our first derivative test.

01:46 So 2 .5 on the left is zero.

01:49 We have a zero in for s prime of t.

01:51 That's going to give us a positive.

01:53 Then if we put three in for s prime of t, we're going to get a negative.

01:57 So we're going positive to negative, which is in fact a maximum.

02:01 So this is a good value for us to use.

02:05 And we just want to check this just to make sure that it is a maximum and not a minimum.

02:10 Of course, this would reach a maximum at some point if it's going up and coming back down.

02:16 So now that we have our t value, we're going to take s of 2 .5, which is going to be 24 .5 times 2 .5...

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