In springboard diving, the diver strides out to the end of...

In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-like nature of the board to help propel him into the air. Assume that the diver's motion is essentially vertical. He leaves the board, which is $3.0 \mathrm{m}$ above the water, with a speed of $6.3 \mathrm{m} / \mathrm{s}$ a. How long is the diver in the air, from the moment he leaves the board until he reaches the water? b. What is the speed of the diver when he reaches the water?

In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-like nature of the board to help propel him into the air. Assume that the diver's motion is essentially vertical. He leaves the board, which is $3.0 \mathrm{m}$ above the water, with a speed of $6.3 \mathrm{m} / \mathrm{s}$
a. How long is the diver in the air, from the moment he leaves the board until he reaches the water?
b. What is the speed of the diver when he reaches the water?

Key Concepts

Kinematic Equations for Uniform Acceleration

This concept involves using the equations of motion that describe the behavior of objects moving with constant acceleration. In problems like this, the vertical displacement, initial velocity, time, and acceleration due to gravity are related through equations such as y = y? + v?t + ½at² and v = v? + at. These equations are essential to analyze the motion of any object in free fall or projectile motion, determining key quantities like time of flight and final velocity.

Projectile Motion (Vertical Component)

Projectile motion refers to the motion of an object thrown or projected into the air, subject only to the acceleration due to gravity. In the vertical component of projectile motion—where the motion is treated as one-dimensional—it's assumed that air resistance is negligible and that the only acceleration acting on the object is gravity. This allows for the analysis of time of flight, maximum height, and impact speed through the kinematic equations.

Application of the Quadratic Formula in Kinematics

When solving kinematics problems, especially those involving displacement over time under constant acceleration, one often ends up with a quadratic equation. The quadratic formula is used to find the values of time at which certain conditions are met (for example, when an object reaches a particular height). This technique is crucial in accurately determining the time intervals during which various phases of the motion occur, such as the duration of free fall.

Transcript

00:01 In this problem, we're going to talk about gravitational acceleration.

00:03 So remember that the gravitational acceleration g on earth is equal to 9 .8 meters per second squared.

00:12 Also, remember that if we set up a coordinate system such that the y -axis is pointing upwards, and hence the gravitational acceleration is pointing downwards, then y is equal to y0, why is the position of the particle, plus the initial velocity be zero times t minus g t squared over two where the minus sign is here because g is pointing down and what we have in our exercise is a diver that jumps as shown here in the picture from a board with an initial velocity of 6 .3 meters per second the board is at a height h that is equal to three meters above the water and we want to calculate in question a how long it takes for the diver to hit the water so he makes this trajectory that's shown in the dashed red line and he hits the water after time t, we want to calculate this time t.

01:38 I'm going to choose the origin of my coordinates of my y -axis to be on the surface of the water, meaning that the initial position, y -0 of the diver, is 3 meters.

01:50 And i'm going to substitute in the equation for the y coordinates.

01:53 We have that y is equal to 3 meters plus 6 .3 meters per second times the time t, minus g divided by 2, that's 9 .8 divided by 2 or 4 .9 meter per second squared times t squared.

02:14 And we want to calculate t when the diver hits the water, meaning that y is equal to 0.

02:21 So 0 is equal to 3 plus 6 .3 t minus 4 .9 t squared.

02:30 I'm going to just write this as 4 .9 t squared minus 6 .9 t squared minus 6 .3 t minus three.

02:42 So i'm going to use basquare to solve this...

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