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A diver jumps from a \(40.0 \mathrm{~m}\) high cliff into the sea. Rocks stick out of the water for a horizontal distance of \(7.00 \mathrm{~m}\) from the foot of the cliff. With what minimum horizontal speed must the diver jump off the cliff in order to clear the rocks and land safely in the sea?

Short Answer

Expert verified
Answer: The minimum horizontal speed required for the diver to clear the rocks and land safely is approximately 2.44 m/s.

Step by step solution

01

Calculate time taken in vertical motion

To calculate the time taken for the diver to fall 40 meters, we can use the following kinematic equation: \(h = ut + \frac{1}{2}gt^2\) where \(h\) is the vertical distance fallen (40 meters), \(u\) is the initial vertical velocity (0, since the diver jumps off horizontally), \(g\) is the acceleration due to gravity (\(9.81 \mathrm{m/s^2}\)), and \(t\) is the time taken. Plugging in the values, we get: \(40 = 0 \cdot t + \frac{1}{2}(9.81)t^2\)
02

Solve for time

We can solve this equation for \(t\), the time taken for the diver to fall 40 meters: \(40 = \frac{1}{2}(9.81)t^2\) To find the value of \(t\), we can first multiply both sides of the equation by 2: \(80 = (9.81)t^2\) Now dividing both sides by 9.81: \(t^2 = \frac{80}{9.81}\) Taking the square root: \(t = \sqrt{\frac{80}{9.81}}\) Using a calculator, we find that: \(t \approx 2.87 \mathrm{s}\)
03

Calculate required horizontal speed

Now that we have the time taken for the diver to fall 40 meters, we can find the required horizontal speed to clear the 7 meters of rocks. We'll use the following equation for horizontal motion: \(d = vt\) where \(d\) is the horizontal distance (7 meters), \(v\) is the horizontal speed, and \(t\) is the time calculated in step 2 (\(2.87 \mathrm{s}\)). Plugging in the values: \(7 = v(2.87)\) Now, we can solve for \(v\): \(v = \frac{7}{2.87}\) Using a calculator, we find the required horizontal speed: \(v \approx 2.44 \mathrm{m/s}\) The diver must jump off the cliff with a minimum horizontal speed of \(2.44 \mathrm{m/s}\) to clear the rocks and land safely in the sea.

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Key Concepts

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

Horizontal Projectile Motion
When a diver jumps off a cliff and moves horizontally, this type of motion is called horizontal projectile motion. In this motion, there are two components at play: horizontal and vertical. The horizontal component deals with the horizontal distance (in this case, 7 meters of rocks), while the vertical component deals with the fall (from the 40-meter-high cliff).
  • **Horizontal Component**: The horizontal velocity remains constant throughout the dive since there are no horizontal forces acting on the diver once in the air (assuming air resistance is negligible).
  • **Vertical Component**: The vertical motion is influenced by gravity, which accelerates the diver downward.
Understanding these two components is crucial in solving problems related to projectile motion, like figuring out the speed needed to clear obstacles safely.
Kinematic Equations
Kinematic equations are powerful tools to solve problems involving motion, such as the divers leap from the cliff. These equations relate the five key variables: initial velocity, final velocity, acceleration, time, and displacement. In this scenario, we used two main kinematic equations.
  • The first is for vertical motion: \(h = ut + \frac{1}{2}gt^2\), where \(h\) is height, \(u\) is initial vertical velocity, \(g\) is gravity, and \(t\) is time. Here, \(u = 0\) because the leap is horizontal, so the diver starts with no vertical velocity.
  • The second is for horizontal motion: \(d = vt\), where \(d\) is the horizontal distance traveled, \(v\) is horizontal speed, and \(t\) is the time.
These equations enable us to calculate how long it takes the diver to fall and the speed needed to clear the horizontal distance.
Acceleration Due to Gravity
Gravity is a fundamental force acting on objects in free fall, like our diver. On Earth, its constant value is approximately \(9.81 \mathrm{m/s^2}\). This means that a freely falling object increases its velocity by \(9.81 \mathrm{m/s}\) every second.
  • In the diver's case, gravity is what pulls them downward, causing the vertical component of their jump.
  • It affects only the vertical motion, not the horizontal.
By knowing the gravitational acceleration, we can determine how long it takes to fall a certain height, critical for finding how fast the diver should move horizontally to avoid obstacles.

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Most popular questions from this chapter

A projectile is launched at an angle of \(45^{\circ}\) above the horizontal. What is the ratio of its horizontal range to its maximum height? How does the answer change if the initial speed of the projectile is doubled?

For a science fair competition, a group of high school students build a kicker-machine that can launch a golf ball from the origin with a velocity of \(11.2 \mathrm{~m} / \mathrm{s}\) and initial angle of \(31.5^{\circ}\) with respect to the horizontal. a) Where will the golf ball fall back to the ground? b) How high will it be at the highest point of its trajectory? c) What is the ball's velocity vector (in Cartesian components) at the highest point of its trajectory? d) What is the ball's acceleration vector (in Cartesian components) at the highest point of its trajectory?

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A baseball is launched from the bat at an angle \(\theta_{0}=30^{\circ}\) with respect to the positive \(x\) -axis and with an initial speed of \(40 \mathrm{~m} / \mathrm{s}\), and it is caught at the same height from which it was hit. Assuming ideal projectile motion (positive \(y\) -axis upward), the velocity of the ball when it is caught is a) \((20.00 \hat{x}+34.64 \hat{y}) \mathrm{m} / \mathrm{s}\). b) \((-20.00 \hat{x}+34.64 \hat{y}) \mathrm{m} / \mathrm{s}\) c) \((34.64 \hat{x}-20.00 \hat{y}) \mathrm{m} / \mathrm{s}\) d) \((34.64 \hat{x}+20.00 \hat{y}) \mathrm{m} / \mathrm{s}\).

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