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Chapter 3: Problem 71

Multiple-Concept Example 4 provides useful background for this problem. A diver runs horizontally with a speed of 1.20 m/s off a platform that is 10.0 m above the water. What is his speed just before striking the water?

Short Answer

Expert verified
The diver's speed just before hitting the water is approximately 14.01 m/s.

Step by step solution

01

Analyze the Given Information

The diver runs horizontally off a platform, so the initial vertical velocity is 0 m/s. The horizontal velocity when he leaves the platform is 1.20 m/s, and the height from which he jumps is 10.0 m.
02

Determine Time to Fall

Use the formula for free fall to calculate the time it takes to fall 10.0 m: \[ h = \frac{1}{2} g t^2 \] where \( h = 10.0 \) m and \( g = 9.81 \text{ m/s}^2 \). Solving for \( t \): \[ 10.0 = \frac{1}{2} \times 9.81 \times t^2 \] \[ t^2 = \frac{10.0}{4.905} \] \[ t = \sqrt{2.03866} \approx 1.43 \text{ s} \]
03

Calculate Vertical Speed Before Impact

Calculate the vertical speed just before hitting the water using the formula: \[ v = g \times t \] where \( g = 9.81 \text{ m/s}^2 \) and \( t = 1.43 \text{ s} \): \[ v = 9.81 \times 1.43 \approx 14.03 \text{ m/s} \]
04

Determine Resultant Speed Just Before Impact

Find the resultant speed using the Pythagorean theorem, since the motion involves independent horizontal and vertical components:\[ v_{\text{resultant}} = \sqrt{v_x^2 + v_y^2} \] where \( v_x = 1.20 \text{ m/s} \) and \( v_y = 14.03 \text{ m/s} \): \[ v_{\text{resultant}} = \sqrt{(1.20)^2 + (14.03)^2} \approx \sqrt{196.4089} \approx 14.01 \text{ m/s} \]
05

Conclusion

The final speed of the diver just before striking the water, taking into account both the horizontal and the vertical components of the velocity, is approximately 14.01 m/s.

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Key Concepts

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

Horizontal Velocity
Understanding horizontal velocity is crucial in projectile motion. It refers to the constant speed at which an object moves horizontally. For example, when a diver leaps off a platform, their horizontal velocity remains unchanged because there is no horizontal force acting on them. This is a classic case of Newton's first law of motion, which tells us that an object in motion will stay in motion at a constant velocity unless acted upon by another force. In our problem, the diver's horizontal velocity is 1.20 m/s. Even as the diver falls, this speed does not change, demonstrating the independence of horizontal motion from vertical influences. This concept helps us predict the horizontal distance the diver will cover before hitting the water.
Vertical Velocity
Vertical velocity is different from horizontal since it involves movement under the influence of gravity. Initially, when our diver steps off the platform, their vertical velocity is zero. However, as they fall, gravity accelerates them downwards at 9.81 m/s虏. This gravitational pull increases the diver's vertical speed over time. After 1.43 seconds of free fall, the diver's vertical velocity just before hitting the water is approximately 14.03 m/s. Understanding this concept is key to determining the vertical distance and speed at different time points during the dive.
Free Fall
Free fall describes the motion of an object solely under the influence of gravity, drawing it constantly downward. Our diver, once off the platform, is in free fall. Thanks to gravity, his speed increases until he reaches the water. The time taken for the diver to fall 10 meters is found using the equation for free fall: \[ h = \frac{1}{2} g t^2 \]. By solving this equation, we calculate the time of 1.43 seconds it takes for the diver to reach the water. Recognizing free fall helps predict how fast and when the diver will hit the ground.
Pythagorean Theorem
The Pythagorean Theorem provides a way to find the resultant velocity when dealing with both horizontal and vertical components of motion. Since these components are perpendicular, they form a right triangle. In this case, the horizontal motion (1.20 m/s) and vertical motion (14.03 m/s) are the two legs of a right triangle. Using the Pythagorean theorem: \[ v_{\text{resultant}} = \sqrt{v_x^2 + v_y^2} \], we find the hypotenuse, or the resultant speed, is approximately 14.01 m/s. This approach illustrates how different components of motion combine to determine overall speed.

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Most popular questions from this chapter

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