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

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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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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.
Transcribed Image Text:**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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