VIPSolutionsgive answer in 2 step with explanation at the end of each step and final answer at the end:24. An arrow is shot into the air. Its vertical component is 80 km/h and the magnitude of its initial velocity is 95 km/h. Determine e the the horizontal horizontal component of the arrow at the moment ment it it is is released. (2 marks)
Question:
give answer in 2 step with explanation at the end of each step and final answer at the end:24. An arrow is shot into the air. Its vertical component is 80 km/h and the magnitude of its initial velocity is 95 km/h. Determine e the the horizontal horizontal component of the arrow at the moment ment it it is is released. (2 marks)
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Asked by: Rahul
Created at: 2026-02-06 04:52:19
Asked by: Rahul
Created at: 2026-02-06 04:52:19
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# Step 1: Use the Pythagorean Theorem
Given:
- Vertical component (\(v_y\)) = 80 km/h
- Magnitude of initial velocity (\(v\)) = 95 km/h
The components form a right triangle:
\[
v = \sqrt{v_x^2 + v_y^2}
\]
where \(v_x\) is the horizontal component.
Rearrange to solve for \(v_x\):
\[
v_x = \sqrt{v^2 - v_y^2}
\]
**Explanation:**
The initial velocity vector and its vertical and horizontal components form a right triangle, so we use the Pythagorean theorem.
---
# Step 2: Substitute Values and Solve
\[
v_x = \sqrt{(95)^2 - (80)^2}
\]
\[
v_x = \sqrt{9025 - 640}
\]
\[
v_x = \sqrt{2625}
\]
\[
v_x \approx 51.24 \text{ km/h}
\]
**Explanation:**
After substituting the values and calculating, we find the horizontal component.
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# Final Answer
**The horizontal component of the arrow's velocity at release is approximately \(51.2\) km/h.**
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