**Projectile Motion Analysis: Time to Reach Maximum Height****Problem Statement:**A projectile is launched with a speed \(v_0 = 50 \, \text{m/s}\) at an angle \(\theta = 35^\circ\). How long does it take to reach the maximum height?**Diagram Description:**The diagram shows a projectile being launched from the origin. The initial velocity vector \(v_0\) forms an angle \(\theta\) with the horizontal axis \(x\). The vector is indicated with a red arrow, and the launch angle \(\theta = 35^\circ\) is marked. The vertical axis is labeled \(y\).**Multiple Choice Answers:**- \(\boldsymbol{2.55 \, \text{s}}\)- \(\boldsymbol{28.68 \, \text{s}}\)- \(\boldsymbol{2.92 \, \text{s}}\)- \(\boldsymbol{5.10 \, \text{s}}\)**Solution Approach:**To determine the time to reach the maximum height, we use the following kinematic equation for vertical motion:\[ v_{y} = v_0 \sin(\theta) - g t \]At the maximum height, the vertical component of the velocity \(v_{y}\) is zero:\[ 0 = v_0 \sin(\theta) - g t \]Solving for \(t\):\[ t = \frac{v_0 \sin(\theta)}{g} \]Given:\( v_0 = 50 \, \text{m/s} \)\( \theta = 35^\circ \)\( g \approx 9.8 \, \text{m/s}^2 \)Calculate:\[ t = \frac{50 \sin(35^\circ)}{9.8} \approx \frac{50 \times 0.5736}{9.8} \approx \frac{28.68}{9.8} \approx 2.93 \, \text{s} \]**Correct Answer:** \(\boldsymbol{2.92 \, \text{s}}\)

Introduction: A projectile is any item hurled into space with just gravity acting on it. Gravity is…

Answered: A projectile is launched with a speed…

A projectile is launched with a speed v0 = 50 m/s at angle = 35°. How long does it take to reach the maximum height? y vo 0 2.55 s 28.68 s O2.92 s 5.10 S

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Basic Terminology in Mechanics

The area of science and mathematics concerned the motion of the objects, the relation between the force matter, and motion if specified. Forces exerted over the objects result in displacements or changes in the position of an object concerning its environment. This branch of physics keeps its roots in Ancient Greece in the form of the scripts of Aristotle and Archimedes.

Question

**Projectile Motion Analysis: Time to Reach Maximum Height**

**Problem Statement:**
A projectile is launched with a speed \(v_0 = 50 \, \text{m/s}\) at an angle \(\theta = 35^\circ\). How long does it take to reach the maximum height?

**Diagram Description:**
The diagram shows a projectile being launched from the origin. The initial velocity vector \(v_0\) forms an angle \(\theta\) with the horizontal axis \(x\). The vector is indicated with a red arrow, and the launch angle \(\theta = 35^\circ\) is marked. The vertical axis is labeled \(y\).

**Multiple Choice Answers:**
- \(\boldsymbol{2.55 \, \text{s}}\)
- \(\boldsymbol{28.68 \, \text{s}}\)
- \(\boldsymbol{2.92 \, \text{s}}\)
- \(\boldsymbol{5.10 \, \text{s}}\)

**Solution Approach:**
To determine the time to reach the maximum height, we use the following kinematic equation for vertical motion:
\[ v_{y} = v_0 \sin(\theta) - g t \]

At the maximum height, the vertical component of the velocity \(v_{y}\) is zero:
\[ 0 = v_0 \sin(\theta) - g t \]

Solving for \(t\):
\[ t = \frac{v_0 \sin(\theta)}{g} \]

Given:
\( v_0 = 50 \, \text{m/s} \)
\( \theta = 35^\circ \)
\( g \approx 9.8 \, \text{m/s}^2 \)

Calculate:
\[ t = \frac{50 \sin(35^\circ)}{9.8} \approx \frac{50 \times 0.5736}{9.8} \approx \frac{28.68}{9.8} \approx 2.93 \, \text{s} \]

**Correct Answer:** \(\boldsymbol{2.92 \, \text{s}}\)

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