A ball is thrown vertically upward from a height of 6 feet with an initial velocity of 60 feet p

A ball is thrown vertically upward from a height of 6 feet...

Key Concepts

Kinematic Equations for Uniformly Accelerated Motion

These equations relate the displacement, initial velocity, acceleration, and time for an object moving with constant acceleration. They provide a systematic approach to solving problems in projectile motion, such as finding the time it takes for an object to reach its maximum height or its maximum vertical displacement.

Acceleration Due to Gravity

This concept refers to the constant acceleration experienced by objects in free fall near the Earth's surface. It acts downward and is a crucial factor when analyzing vertically thrown objects, as it decelerates the object until its upward velocity becomes zero, and then accelerates it downward.

Maximum Height in Projectile Motion

The maximum height is reached when the vertical velocity of the projectile becomes zero. By applying the kinematic equations and setting the final vertical velocity at the peak to zero, one can determine the highest point the object reaches relative to its initial position. This concept is central to problems involving objects thrown vertically upward.

Transcript

00:01 So in this question, we are first told that the acceleration of an object due to gravity is given by negative 32 feet per second.

00:11 Per second.

00:12 Okay.

00:13 So now, we are given that a ball is thrown vertically upward from a height of six feet with an initial velocity of 60 feet per second.

00:24 So what have we been given there? we've been given, first of all, that my velocity at time zero is, 60 feet per second, and i'm told that we have an initial height of six feet.

00:39 So the height of this object at time zero is six feet.

00:44 So my initial velocity is 60.

00:47 My initial height is six.

00:49 And i want to eventually figure out the maximum height of this object.

00:56 Okay.

00:58 So how are we going to do that? well, i was given acceleration, yes? and i know that velocity.

01:04 Is the anti -derivative of acceleration.

01:09 Yes? if acceleration is the derivative of velocity, velocity is the anti -derivative of my acceleration.

01:20 And so my velocity this time is the anti -derivative of my acceleration.

01:25 It's the anti -derivative of negative 32 dt.

01:31 What's the anti -derivative of negative 32 with respect to t? well, that's just going to be negative 32t.

01:43 Don't forget to add your constant c.

01:47 Now, i can figure out that constant c since i know that my velocity at times zero is 60.

01:55 So my velocity at time zero by plugging in, i'm getting zero plus c.

02:05 V of zero is 60.

02:08 And so my 60, that's my c value.

02:13 And so i'll take that c.

02:17 I'll plug it in.

02:19 And so what i'm getting here is that i have a velocity function, v of t equals negative 32t plus 60.

02:30 Now from here, i'm going to go back and get my height function.

02:36 Once i have velocity, i know my height, which is my position function, is the anti -derivative of what? my velocity function.

02:48 Velocity is the derivative of position.

02:51 So position is the anti -derivative of velocity.

02:57 And so my height here is going to be the anti -derivative of negative 32t plus 60 with respect to t...

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