Problem 37 - Fundamentals of Physics

Question

A lowly high diver pushes off horizontally with a speed of \(2.00 \mathrm{~m} / \mathrm{s}\) from the platform edge \(10.0 \mathrm{~m}\) above the surface of the water. (a) At what horizontal distance from the edge is the diver \(0.800 \mathrm{~s}\) after pushing off? (b) At what vertical distance above the surface of the water is the diver just then? (c) At what horizontal distance from the edge does the diver strike the water?

Step-by-Step Solution

Answer

(a) 1.60 m, (b) 6.86 m, (c) 2.86 m.

Step 1: Understand the problem

A diver pushes off from a platform horizontally with an initial horizontal speed and no initial vertical velocity. She falls under gravity and we need to find her horizontal position and vertical fall after certain time intervals, and when she hits the water surface.

Step 2: Horizontal motion calculation

Since there is no horizontal acceleration, the horizontal distance covered in time is given by \(x = v_x \cdot t\) where \(v_x = 2.00 \ \mathrm{m/s}\). After \(0.800\ \mathrm{s}\), the horizontal distance is \(x = 2.00 \cdot 0.800 = 1.60 \ \mathrm{m}\).

Step 3: Vertical motion calculation

The vertical distance is found using the equation for motion under gravity: \(y = v_{y_0} t + \frac{1}{2} g t^2\), where \(v_{y_0} = 0\ \mathrm{m/s}\) since the motion was purely horizontal and \(g = 9.81 \ \mathrm{m/s^2}\). Thus, \(y = 0 + 0.5 \cdot 9.81 \cdot (0.800)^2 = 3.14 \ \mathrm{m}\). The diver is \(10.0 - 3.14 = 6.86\ \mathrm{m}\) above the water after \(0.800\ \mathrm{s}\).

Step 4: Time to hit the water

To find when the diver hits the water, set \(y = 10.0\ \mathrm{m}\): \(10.0 = 0 + 0.5 \cdot 9.81 \cdot t^2\). Solving for \(t\) gives \(t^2 = \frac{20.0}{9.81}\), leading to \(t \approx 1.43\ \mathrm{s}\).

Step 5: Horizontal distance when hitting water

Using the time calculated, the horizontal distance when the diver hits the water is \(x = v_x \cdot t = 2.00 \cdot 1.43 = 2.86\ \mathrm{m}\).

Key Concepts

Horizontal Motion

When we discuss horizontal motion in physics, we're talking about how objects move in a straight line across a flat plane. In the case of our diver, this motion is straightforward because she begins with a horizontal speed of \(2.00 \,\text{m/s}\). Since there is no horizontal force acting on her after she leaves the platform, she will continue moving at this constant speed.

This type of motion follows the simple equation \(x = v_x \cdot t\), where \(x\) is the horizontal distance, \(v_x\) is the horizontal velocity, and \(t\) is the time elapsed. This means:

• There is no change in speed — it's constant.
• The movement is unimpeded and smooth.
• It follows a straight line with no deviation.

For our diver, after \(0.800 \,\text{s}\), her horizontal position will be \(1.60 \,\text{m}\) from the starting point, as calculated from \(2.00 \cdot 0.800\).

Vertical Motion

Vertical motion is a bit more complex than horizontal motion since gravity comes into play here. When the diver leaps off the platform, she begins to fall under the influence of gravity, which accelerates her downwards at \(9.81 \,\text{m/s}^2\).

Initially, our diver has no vertical velocity (\(v_{y_0} = 0\)) because she jumps off horizontally. The vertical distance she falls over time \(t\) can be found using the equation: \(y = v_{y_0}t + \frac{1}{2} g t^2\). This tells us that:

• She starts with no downward speed, just like how something dropped straight down.
• Gravity progressively increases her speed as she falls.
• Her descent follows a parabolic path due to constant acceleration.

Employing the equation for our diver, after \(0.800 \,\text{s}\), we've calculated her vertical fall to be \(3.14 \,\text{m}\), leaving her \(6.86 \,\text{m}\) above the water.

Kinematics

Kinematics is the branch of physics that deals with motion itself, breaking down movements into simpler components, like horizontal and vertical motions. Understanding kinematics allows us to predict where an object, like our diver, will be after any given time without considering the forces that cause the motion. It focuses on:

• The measurements of travel (displacements).
• Speeds and how they change over time (velocities).
• The rate of change of speeds (accelerations).

In our diver's leap, kinematic equations help us separately analyze her horizontal travel and vertical fall, combining these outcomes to pinpoint her precise location at various times after leaving the platform. As Kinematics doesn't deal with why an object moves, we focus just on the descriptive aspects of motion.

Gravity

Gravity is the invisible force that heavily influences vertical motion in projectile movements. It pulls all objects downwards towards the earth at approximately \(9.81 \,\text{m/s}^2\). This constant acceleration due to gravity acts immediately upon any object that is in free fall, such as our diver.

Gravity plays a crucial role:

• It ensures that, as soon as the diver leaves the horizontal plane, she begins to drop towards the ground.
• The acceleration remains constant, which is why the kinematic equations include \(\frac{1}{2} g t^2\) for vertical motion.
• Even with forward motion, gravity is still effective downward.

Knowing gravity's acceleration value allows us to predict how quickly an object will change vertically over time, as seen in our diver's descent towards the water.

Free Fall

Free fall describes a situation where gravity is the only significant force acting on an object. In this context, the diver's vertical motion can be considered a free fall since, after she pushes off the platform, gravity solely dictates her vertical pathway.

In free fall, the object:

• Starts from rest with an initial velocity of zero, especially for the vertical component.
• Falls under gravity's constant force without any additional propulsion.
• Accelerates downwards steadily, evident in the increasing vertical distance fallen over time.

Our diver's free fall is the vertical part of her jump. While she retains her horizontal speed, her downward journey is entirely governed by gravity, allowing us to predict the time when she'll reach the surface with precision.