Thank you for visiting A ball is tossed from an upper story window of a building The ball is given an initial velocity of tex 8 00 text m. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Sure, let's solve this problem step by step.
1. Understand the problem:
We need to find how far horizontally from the base of the building a ball lands after being tossed from an upper-story window. The ball has an initial velocity of [tex]\(8.00 \, \text{m/s}\)[/tex] at an angle of [tex]\(20.0^\circ\)[/tex] below the horizontal and hits the ground [tex]\(3.00\)[/tex] seconds later.
2. Break down the initial velocity:
The ball's initial velocity can be broken into horizontal and vertical components. The horizontal component ([tex]\(v_{0x}\)[/tex]) will remain constant (assuming air resistance is negligible).
- The horizontal component of the initial velocity is given by:
[tex]\[
v_{0x} = v_0 \cos(\theta)
\][/tex]
Where [tex]\(v_0 = 8.00 \, \text{m/s}\)[/tex] and [tex]\(\theta = 20.0^\circ\)[/tex].
3. Find the horizontal component of the velocity:
Substituting the values:
[tex]\[
v_{0x} = 8.00 \cos(20.0^\circ)
\][/tex]
4. Calculate the horizontal distance:
The horizontal distance ([tex]\(x\)[/tex]) is given by:
[tex]\[
x = v_{0x} \times t
\][/tex]
Where [tex]\(t = 3.00 \, \text{s}\)[/tex].
5. Numerical calculation:
Evaluating the trigonometric function and multiplying:
[tex]\[
\cos(20.0^\circ) \approx 0.9397
\][/tex]
[tex]\[
v_{0x} = 8.00 \times 0.9397 \approx 7.52 \, \text{m/s}
\][/tex]
[tex]\[
x = 7.52 \, \text{m/s} \times 3.00 \, \text{s} \approx 22.56 \, \text{m}
\][/tex]
From the calculations, we find that the horizontal distance from the base of the building where the ball strikes the ground is [tex]\(22.6 \, \text{m}\)[/tex].
Thus, the correct answer is:
A) 22.6 m
1. Understand the problem:
We need to find how far horizontally from the base of the building a ball lands after being tossed from an upper-story window. The ball has an initial velocity of [tex]\(8.00 \, \text{m/s}\)[/tex] at an angle of [tex]\(20.0^\circ\)[/tex] below the horizontal and hits the ground [tex]\(3.00\)[/tex] seconds later.
2. Break down the initial velocity:
The ball's initial velocity can be broken into horizontal and vertical components. The horizontal component ([tex]\(v_{0x}\)[/tex]) will remain constant (assuming air resistance is negligible).
- The horizontal component of the initial velocity is given by:
[tex]\[
v_{0x} = v_0 \cos(\theta)
\][/tex]
Where [tex]\(v_0 = 8.00 \, \text{m/s}\)[/tex] and [tex]\(\theta = 20.0^\circ\)[/tex].
3. Find the horizontal component of the velocity:
Substituting the values:
[tex]\[
v_{0x} = 8.00 \cos(20.0^\circ)
\][/tex]
4. Calculate the horizontal distance:
The horizontal distance ([tex]\(x\)[/tex]) is given by:
[tex]\[
x = v_{0x} \times t
\][/tex]
Where [tex]\(t = 3.00 \, \text{s}\)[/tex].
5. Numerical calculation:
Evaluating the trigonometric function and multiplying:
[tex]\[
\cos(20.0^\circ) \approx 0.9397
\][/tex]
[tex]\[
v_{0x} = 8.00 \times 0.9397 \approx 7.52 \, \text{m/s}
\][/tex]
[tex]\[
x = 7.52 \, \text{m/s} \times 3.00 \, \text{s} \approx 22.56 \, \text{m}
\][/tex]
From the calculations, we find that the horizontal distance from the base of the building where the ball strikes the ground is [tex]\(22.6 \, \text{m}\)[/tex].
Thus, the correct answer is:
A) 22.6 m
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