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Chapter 3: Problem 29

A diver running \(2.3 \mathrm{~m} / \mathrm{s}\) dives out horizontally from the edge of a vertical cliff and \(3.0 \mathrm{~s}\) later reaches the water below. How high was the cliff and how far from its base did the diver hit the water?

Short Answer

Expert verified
The cliff is approximately 44.1 meters high, and the diver hit the water 6.9 meters from the base.

Step by step solution

01

Determine Vertical Motion of the Diver

To find the height of the cliff, we first analyze the vertical motion. The only force acting on the diver vertically is gravity. We can use the equation for distance under constant acceleration: \[ h = rac{1}{2} g t^2 \]where \(g = 9.8 \mathrm{~m/s^2}\) is the acceleration due to gravity and \(t = 3.0 \mathrm{~s}\) is the time.Substituting the values, we get:\[ h = \frac{1}{2} \times 9.8 \times (3.0)^2 \approx 44.1 \mathrm{~m} \]
02

Calculate Horizontal Distance Traveled

Next, we calculate how far the diver travelled horizontally. The diver's horizontal velocity is constant at \(2.3 \mathrm{~m/s}\). The distance can be found using:\[ d = v imes t \]where \(v = 2.3 \mathrm{~m/s}\) and \(t = 3.0 \mathrm{~s}\).Plugging in the values, we find:\[ d = 2.3 \times 3.0 = 6.9 \mathrm{~m} \]

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Key Concepts

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

Vertical Motion
When we talk about vertical motion in the context of projectile motion, we're focusing on an object's movement under the influence of gravity. In our exercise, once the diver leaves the cliff, gravity is the only force acting upon them vertically.

The diver's vertical acceleration is constant and equal to the acceleration due to gravity, denoted as \(g\), which is approximately \(9.8 \text{ m/s}^2\). Because of this constant acceleration, we use the formula:\[ h = \frac{1}{2} gt^2 \]

**Key Considerations:**
  • Gravity causes the diver to accelerate downwards as soon as they leave the cliff.
  • Vertical motion doesn't influence horizontal motion and vice versa; they are independent in projectile motion.
  • Time \(t\) is crucial in this calculation as it determines how long gravity acts on the diver.

By plugging in the values \(g = 9.8 \text{ m/s}^2\) and \(t = 3.0 \text{ s}\), we find the height of the cliff: \(44.1 \text{ m}\).
Horizontal Motion
Horizontal motion in projectile scenarios like this is quite straightforward. The diver's initial forward velocity of \(2.3 \text{ m/s}\) remains unchanged throughout the dive. This is because there are generally no horizontal forces (like air resistance, unless specified) acting on the diver.

Use the equation for horizontal distance, which is simply speed multiplied by time:\[ d = vt \]

**Key Points to Remember:**
  • The horizontal speed remains constant.
  • Horizontal distance depends entirely on the initial speed and how long the diver stays in the air.
  • Time for both vertical and horizontal motions is always the same, as both occur simultaneously.

For our problem, since \(v = 2.3 \text{ m/s}\) and \(t = 3.0 \text{ s}\), the diver lands \(6.9 \text{ m}\) away from the base of the cliff.
Constant Acceleration
The term "constant acceleration" refers to scenarios where the velocity of an object increases at a steady rate over time. In our diver's case, gravity exerts a constant acceleration on them throughout their flight until they hit the water.

This concept is crucial because it allows us to predict vertical motion reliably using well-established kinematic equations.

Some things to keep in mind about constant acceleration include:
  • Gravity鈥檚 acceleration \(g\) is considered constant at \(9.8 \text{ m/s}^2\), enabling precise calculations.
  • It results in predictable and repeatable paths when objects are projected or dropped from rest.
  • In the absence of other forces, such as air resistance, path computation becomes straightforward and very accurate.

This principle underpins much of classical mechanics and is a foundation for more complex physics studies.

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Most popular questions from this chapter

A particle has a velocity of \(\vec{\mathbf{v}}=(-2.0 \hat{\mathbf{i}}+3.5 t \mathbf{j}) \mathrm{m} / \mathrm{s}\) . The particle starts at \(\vec{\mathbf{r}}=(1.5 \hat{\mathrm{i}}-3.1 \hat{\mathrm{j}}) \mathrm{m}\) at \(t=0 .\) Give the position and acceleration as a function of time. What is the shape of the resulting path?

Extreme-sports enthusiasts have been known to jump off the top of El Capitan, a sheer granite cliff of height \(910 \mathrm{~m}\) in Yosemite National Park. Assume a jumper runs horizontally off the top of El Capitan with speed \(5.0 \mathrm{~m} / \mathrm{s}\) and enjoys a freefall until she is \(150 \mathrm{~m}\) above the valley floor, at which time she opens her parachute (Fig. \(3-41\) ). (a) How long is the jumper in freefall? Ignore air resistance. (b) It is important to be as far away from the cliff as possible before opening the parachute. How far from the cliff is this jumper when she opens her chute?

Huck Finn walks at a speed of \(0.70 \mathrm{~m} / \mathrm{s}\) across his raft (that is, he walks perpendicular to the raft's motion relative to the shore). The raft is traveling down the Mississippi River at a speed of \(1.50 \mathrm{~m} / \mathrm{s}\) relative to the river bank (Fig. \(3-49\) ). What is Huck's velocity (speed and direction) relative to the river bank?

When Babe Ruth hit a homer over the 8.0 -m-high rightfield fence \(98 \mathrm{~m}\) from home plate, roughly what was the minimum speed of the ball when it left the bat? Assume the ball was hit \(1.0 \mathrm{~m}\) above the ground and its path initially made a \(36^{\circ}\) angle with the ground.

Derive a formula for the horizontal range \(R,\) of a projectile when it lands at a height \(h\) above its initial point. (For \(h<0\), it lands a distance \(-h\) below the starting point.) Assume it is projected at an angle \(\theta_{0}\) with initial speed \(v_{0}\).
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