A ball is thrown directly upward from a height of 32 feet...

A ball is thrown directly upward from a height of 32 feet above the ground with an initial velocity of 80 feet per second. Find the time interval during which the ball will be more than 96 feet above the ground. (Hint: See Exercise 91.)
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Key Concepts

Projectile Motion

Projectile motion refers to the motion of an object thrown or projected into the air, subject only to acceleration due to gravity. This concept involves analyzing the path taken by the object in the vertical (and sometimes horizontal) direction, where the only significant force acting on it is gravity, causing a uniformly accelerated motion downward. It provides the framework for understanding how initial velocity, launch height, and gravitational force interact over time.

Kinematic Equations

Kinematic equations describe the relationships between displacement, velocity, acceleration, and time. In vertical motion under constant acceleration, these equations are used to model the height of an object over time. With an initial height and velocity provided, one of the standard equations, h = h0 + v0*t - (1/2)g*t^2, relates these quantities and allows for the analysis of the object's position at any given time.

Quadratic Equations and Inequalities

When solving for the time interval during which the object reaches a certain height, the kinematic equation becomes a quadratic equation in terms of time. Solving the corresponding quadratic inequality (e.g., finding when the height is greater than a specified value) typically involves finding the roots of the related equation using methods such as the quadratic formula. These roots then help define the time interval during which the inequality holds true.

Transcript

00:01 So in this problem, we're being asked to use the equation from exercise 91, where s is equal to negative 16t squared plus v .0 t plus s sub zero to be able to find the height of this ball.

00:12 Now, in this case, we're told the ball is thrown directly upward from a height 32 feet above the ground.

00:18 So that means s sub zero is 32.

00:20 We're told the initial velocity is 80 feet per second.

00:23 So that's v sub zero.

00:25 And we want to find a time interval when the ball is going to be greater than 96 feet above the ground.

00:30 So s has to be greater than 96.

00:34 So if we substitute these values into this equation, we'd have negative 16t squared plus 80t plus 32, and we want to know when is this greater than 96.

00:46 Okay.

00:47 Well, first thing we're going to do is set this equal to zero.

00:50 So i'm going to subtract 96 from both sides of our equation.

00:53 So that will give us negative 16t squared plus 80t, and then we have 32.

01:01 Minus 96, which is equal to negative 64.

01:06 Now, i notice that each term is divisible by negative 16.

01:09 So i'm going to divide each side by negative 16.

01:12 So that will leave us with t squared minus 5t plus four.

01:17 And don't forget, because we're divided by a negative, we flip our inequality side.

01:21 So less than zero.

01:23 Okay.

01:23 So now what i'm going to do is find my zeros by factoring.