Q4 **Problem:**A cat leaps to catch a bird. If the cat's jump was at 60.0° off the ground and its initial velocity was 5.2 m/s, what is the highest point of its trajectory?**Options:**- ○ 2.07 m- ○ 2.58 m- ○ 1.03 m- ○ 20.8 m**Explanation:**To find the highest point of the trajectory, we can utilize the vertical component of the initial velocity and the formula for the maximum height of a projectile. The vertical velocity is given by:\[ v_{y} = v \cdot \sin(\theta) \]where \( v \) is the initial velocity (5.2 m/s) and \( \theta \) is the angle of projection (60.0°).The maximum height (\( h \)) can then be calculated using the formula:\[ h = \frac{v_{y}^2}{2g} \]where \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).Plug in the values to find the answer.

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Description: Q4 **Problem:**A cat leaps to catch a bird. If the cat's jump was at 60.0° off the ground and its initial velocity was 5.2 m/s, what is the highest point of its trajectory?**Options:**- ○ 2.07 m- ○ 2.58 m- ○ 1.03 m- ○ 20.8 m**Explanation:**To find th

Write the expression for the maximum height. H=u2sin2θ2g Here, u is the initial velocity, θ is the…

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A cat leaps to catch a bird. If the cat's jump was at 60.0° off the ground and its initial velocity was 5.2 m/s, what is the highest point of its trajectory? 2.07 m 2.58 m 1.03 m 20.8 m

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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**Problem:**

A cat leaps to catch a bird. If the cat's jump was at 60.0° off the ground and its initial velocity was 5.2 m/s, what is the highest point of its trajectory?

**Options:**

- ○ 2.07 m
- ○ 2.58 m
- ○ 1.03 m
- ○ 20.8 m

**Explanation:**

To find the highest point of the trajectory, we can utilize the vertical component of the initial velocity and the formula for the maximum height of a projectile. The vertical velocity is given by:

\[ v_{y} = v \cdot \sin(\theta) \]

where \( v \) is the initial velocity (5.2 m/s) and \( \theta \) is the angle of projection (60.0°).

The maximum height (\( h \)) can then be calculated using the formula:

\[ h = \frac{v_{y}^2}{2g} \]

where \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).

Plug in the values to find the answer.

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