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Chapter 4: Problem 37

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?

Short Answer

Expert verified
(a) 1.60 m, (b) 6.864 m, (c) 2.856 m.

Step by step solution

01

Identify Given Data

Here, we have the initial horizontal velocity \( v_x = 2.00 \, \text{m/s} \), the horizontal time \( t = 0.800 \, \text{s} \), and the height of the platform \( h = 10.0 \, \text{m} \). We want to find the horizontal distance after 0.800 s, the vertical distance from the water surface after 0.800 s, and the total horizontal distance when the diver strikes the water.
02

Calculate Horizontal Distance at 0.800 s

The horizontal distance covered by the diver in time \( t \) can be calculated using the formula: \( d_x = v_x \times t \). Here, substituting the given values, \( d_x = 2.00 \, \text{m/s} \times 0.800 \, \text{s} = 1.60 \, \text{m} \).
03

Calculate Vertical Distance at 0.800 s

To find the vertical distance the diver falls in \( 0.800 \, \text{s} \), use the formula for distance under uniform acceleration: \( d_y = \frac{1}{2} \times g \times t^2 \), where \( g = 9.81 \, \text{m/s}^2 \). So, \( d_y = \frac{1}{2} \times 9.81 \, \text{m/s}^2 \times (0.800 \, \text{s})^2 = 0.800 \, \text{m}^2 \). Calculate \( d_y = 3.136 \, \text{m} \). The vertical distance from the water surface is \( 10.0 - 3.136 = 6.864 \, \text{m} \).
04

Calculate Total Time to Reach Water

The total time to fall 10.0 m is derived by rearranging the formula \( h = \frac{1}{2} \, g \, t^2 \) to solve for \( t \). Solve: \( t = \sqrt{\frac{2 \times h}{g}} = \sqrt{\frac{2 \times 10.0}{9.81}} = 1.428 \, \text{s} \).
05

Calculate Total Horizontal Distance

Using the total fall time, calculate the horizontal distance: \( d_{x,\text{ total}} = v_x \times t = 2.00 \, \text{m/s} \times 1.428 \, \text{s} = 2.856 \, \text{m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal Velocity
In projectile motion, one of the most important components to understand is horizontal velocity. It refers to the constant speed at which an object travels horizontally. Unlike vertical motion, horizontal motion doesn鈥檛 accelerate or decelerate.

For a projectile, this velocity remains unchanged if air resistance is neglected. In our problem, the diver pushes off with a horizontal velocity of \(2.00 \, \text{m/s}\). This means that throughout the flight, the diver鈥檚 horizontal speed towards the water remains the same.
  • This consistency allows us to easily calculate horizontal distances using the simple equation \(d_x = v_x \times t\).
  • Since no force acts horizontally, once the diver jumps, they keep moving at \(2.00 \, \text{m/s}\) in the horizontal direction until they hit the water.
Vertical Distance
Vertical distance in projectile motion accounts for the change in height of an object from its starting point. Calculating vertical distance involves gravity since it is the force acting on a projectile vertically.

In the diver's case, we calculate the vertical distance fallen after \(0.800 \, \text{s}\) using the formula:\[d_y = \frac{1}{2} \times g \times t^2\]
  • This formula is derived from the kinematic equations for uniformly accelerated motion.
  • Here, \(g\) is the acceleration due to gravity, and \(t\) is the time after the diver pushes off.
  • The result tells us how far the diver has descended vertically from the 10-meter high platform.
The remaining height above the water was calculated as \(6.864 \, \text{m}\) at \(0.800 \, \text{s}\). This quantity helps us appreciate the diver's path visually.
Acceleration Due to Gravity
The only force acting on a projectile in motion, if air resistance is not considered, is gravity. Acceleration due to gravity on Earth is approximately \(9.81 \, \text{m/s}^2\), which acts downward on any object, including our diver.

This acceleration affects the vertical motion of the projectile. It causes the diver to continuously increase their falling speed until they hit the water.

  • Gravity does not influence horizontal velocity, so it remains constant.
  • Using the acceleration due to gravity, we calculated both the vertical distance fallen and the total time to hit the water.
It is vital to consider this value in all physics problems related to free fall, as it provides a consistent basis for calculating time and distance vertical components.
Kinematics
Kinematics is the branch of mechanics that describes the motion of objects without considering the forces causing the motion. In our exercise, understanding kinematics is essential to solve for distances and times.

Two main kinematic equations were used:
  • The formula \(d_x = v_x \times t\) helped in finding how far the diver travels horizontally.
  • The equation \(h = \frac{1}{2}g t^2\) was used to find the time it takes for the diver to reach the water and to find the vertical distance fallen after a specified time.
These equations express the relationships between different quantities, such as initial velocity, time, distance, and acceleration. By applying kinematics, we can predict the diver's motion through the air accurately. Understanding this allows us to take known values and computationally advance through complex physics problems to find unknowns. This is a key skill in solving physics problems efficiently.

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Most popular questions from this chapter

A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wall to break apart the wall. The machine was not placed near the wall because then arrows could reach it from the castle wall. Instead, it was positioned so that the stone hit the wall during the second half of its flight. Suppose a stone is launched with a speed of \(v_{0}=28.0 \mathrm{~m} / \mathrm{s}\) and at an angle of \(\theta_{0}=40.0^{\circ} .\) What is the speed of the stone if it hits the wall (a) just as it reaches the top of its parabolic path and (b) when it has descended to half that height? (c) As a percentage, how much faster is it moving in part (b) than in part (a)?

You are to throw a ball with a speed of \(12.0 \mathrm{~m} / \mathrm{s}\) at a target that is height \(h=5.00 \mathrm{~m}\) above the level at which you release the ball (Fig. \(4-58\) ). You want the ball's velocity to be horizontal at the instant it reaches the target. (a) At what angle \(\theta\) above the horizontal must you throw the ball? (b) What is the horizontal distance from the release point to the target? (c) What is the speed of the ball just as it reaches the target?

The fast French train known as the TGV (Train 脿 Grande Vitesse) has a scheduled average speed of \(216 \mathrm{~km} / \mathrm{h}\). (a) If the train goes around a curve at that speed and the magnitude of the acceleration experienced by the passengers is to be limited to \(0.050 \mathrm{~g}\), what is the smallest radius of curvature for the track that can be tolerated? (b) At what speed must the train go around a curve with a \(1.00 \mathrm{~km}\) radius to be at the acceleration limit?

A moderate wind accelerates a pebble over a horizontal \(x y\) plane with a constant acceleration \(\vec{a}=\left(5.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(7.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\) . At time \(t=0,\) the velocity is \((4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}} .\) What are the (a) magnitude and (b) angle of its velocity when it has been displaced by \(12.0 \mathrm{~m}\) parallel to the \(x\) axis?

In basketball, hang is an illusion in which a player seems to weaken the gravitational acceleration while in midair. The illusion depends much on a skilled player's ability to rapidly shift the ball between hands during the flight, but it might also be supported by the longer horizontal distance the player travels in the upper part of the jump than in the lower part. If a player jumps with an initial speed of \(v_{0}=7.00 \mathrm{~m} / \mathrm{s}\) at an angle of \(\theta_{0}=35.0^{\circ},\) what percent of the jump's range does the player spend in the upper half of the jump (between maximum height and half maximum height \()\) ?
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