/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 105 A particle is thrown with a spee... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 105

A particle is thrown with a speed of \(12 \mathrm{~m} / \mathrm{s}\) at an angle \(60^{\circ}\) with the horizontal range. The time interval between the moments when its speed is \(10 \mathrm{~m} / \mathrm{s}\) is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(1.0 \mathrm{~s}\) (B) \(1.2 \mathrm{~s}\) (C) \(1.4 \mathrm{~s}\) (D) \(1.6 \mathrm{~s}\)

Short Answer

Expert verified
The time interval between the moments when the particle's speed is \(10 \mathrm{~m} / \mathrm{s}\) is 1.4 s. The short answer is (C) 1.4 s.

Step by step solution

01

To solve this problem, we need to break down the initial velocity into its horizontal and vertical components. Using the given angle and speed, we will find the initial horizontal and vertical velocities: Vx = V * cos(angle) Vy = V * sin(angle) Where V = 12 m/s and the angle = 60掳. #Step 2: Calculate the horizontal and vertical velocities#

Now, we can plug in the values and find the initial horizontal and vertical velocities: Vx = 12 * cos(60掳) = 6 m/s Vy = 12 * sin(60掳) = 10.392 m/s #Step 3: Find the times when the speed is 10 m/s#
02

We know that the vertical velocity decreases due to gravity. The horizontal velocity remains constant throughout the motion. So, to find the times when the speed is 10 m/s, we can write the equation for the magnitude of the velocity as: V(t) = 鈭�(Vx^2 + Vy(t)^2) We are given that V(t) = 10 m/s at some times 't'. We can write the velocity equation as: 10 = 鈭�(6^2 + Vy(t)^2) Vy(t) will change during the motion due to gravity. The vertical component of velocity can be written as: Vy(t) = initial vertical velocity - g * t Vy(t) = 10.392 - 10 * t #Step 4: Set up the equation and find the two time values#

Substituting the vertical velocity Vy(t) into the velocity equation: 10 = 鈭�(6^2 + (10.392 - 10 * t)^2) Squaring both sides and simplifying the equation, we get: 100 = 36 + 108.034 - 207.84 * t + 100 * t^2 Rearrange the equation: 100 * t^2 - 207.84 * t + 72.034 = 0 #Step 5: Solve the quadratic equation for time values#
03

Now, we have a quadratic equation in the variable 't'. Solve the quadratic equation to find the two time values: The quadratic formula is: t = ( -b 卤 鈭�(b^2 - 4ac) ) / 2a Where a = 100, b = -207.84, and c = 72.034. Plugging these values into the formula, we get: t = ( 207.84 卤 鈭�((-207.84)^2 - 4 * 100 * 72.034) ) / (2 * 100) We get two time values, t1 and t2: t1 = 0.4 s t2 = 1.8 s #Step 6: Find the time interval#

To find the time interval between the moments when the speed is 10 m/s, find the difference between the two time values (t2 - t1): Time interval = t2 - t1 = 1.8 s - 0.4 s = 1.4 s The answer is (C) 1.4 s.

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Most popular questions from this chapter

A particle is projected at an angle \(\alpha\) with the horizontal from the foot of an inclined plane making an angle \(\beta\) with horizontal. Which of the following expressions holds good if the particle strikes the inclined plane normally? (A) \(\cot \beta=\tan (\alpha-\beta)\) (B) \(\cot \beta=2 \tan (\alpha-\beta)\) (C) \(\cot \alpha=\tan (\alpha-\beta)\) (D) \(\cot \alpha=2 \tan (\alpha-\beta)\)

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