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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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