/*! 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 188 From the top of a wall of height... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 188

From the top of a wall of height \(5 \mathrm{~m}\), a ball is thrown horizontally with speed of \(6 \mathrm{~m} / \mathrm{s}\). How far from the wall will the ball land?

Short Answer

Expert verified
The ball will land approximately \(6.06 \,\text{m}\) away from the wall.

Step by step solution

01

Identify the givens and unknowns.

In this problem, we know the following information: - Initial height of the ball from the ground: \(5 \,\text{m}\) - Initial speed of the ball: \(u = 6 \,\text{m/s}\) - Vertical acceleration due to gravity: \(g = 9.8 \,\text{m/s}^2\) We are trying to find the horizontal distance (range) traveled by the ball.
02

Calculate the time taken for the ball to fall to the ground.

We can use the equation of motion to determine the time it takes for the ball to fall to the ground. In the vertical direction, the equation of motion is: \(y = y_0 + v_{0y}t - \frac{1}{2}gt^2\) Since the ball is thrown horizontally, it has no initial vertical speed. Therefore, \(v_{0y} = 0\). The final height should also be 0 when the ball lands: \(0 = 5 - \frac{1}{2}(9.8)t^2\) Now, solve for \(t\): \(5 = \frac{1}{2}(9.8)t^2\)
03

Solve for t

Divide both sides by 4.9: \(\frac{5}{4.9} = t^2\) Now, take the square root to find \(t\): \(t = \sqrt{\frac{5}{4.9}} \approx 1.01 \, \text{seconds}\)
04

Calculate the horizontal distance traveled.

Since we have the time it takes for the ball to fall to the ground, we can now calculate the horizontal distance (range) using the horizontal speed of the ball: \(Range = u \times time\) \(Range = (6 \,\text{m/s}) \times (1.01 \,\text{s})\)
05

Find the horizontal distance

Multiply to find the range: \( Range \approx 6.06 \,\text{m}\) The ball will land approximately \(6.06 \,\text{m}\) away from the wall.

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Most popular questions from this chapter

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