⠀→ A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone when it falls back to the ground ?

⠀→ A stone is thrown vertically upward with an initial velocity of

1.	⠀→ A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone when it falls back to the ground ?
Answer» <html><body><p><strong>Answer:</strong></p><ul><li><strong>Maximum</strong><strong> height</strong><strong> attained</strong><strong> by</strong><strong> the</strong><strong> </strong><strong>stone</strong><strong> </strong><strong>=</strong><strong> </strong><strong>8</strong><strong>0</strong><strong>m</strong></li><li><strong><a href="https://interviewquestions.tuteehub.com/tag/net-5194" style="font-weight:bold;" target="_blank" title="Click to know more about NET">NET</a> </strong><strong><a href="https://interviewquestions.tuteehub.com/tag/displacement-956081" style="font-weight:bold;" target="_blank" title="Click to know more about DISPLACEMENT">DISPLACEMENT</a></strong><strong> </strong><strong>=</strong><strong> </strong><strong>0</strong><strong> </strong><strong>m</strong></li><li><strong>Total</strong><strong> </strong><strong>Distance</strong><strong> </strong><strong>covered</strong><strong> by</strong><strong> </strong><strong>stone</strong><strong> </strong><strong>=</strong><strong> </strong><strong>1</strong><strong>6</strong><strong>0</strong><strong> </strong><strong>m </strong></li></ul><p><strong>Explanation:</strong></p><p></p><p><strong><<a href="https://interviewquestions.tuteehub.com/tag/u-1435036" style="font-weight:bold;" target="_blank" title="Click to know more about U">U</a>>Given:</strong><strong><u>-</u></strong></p><p></p><ul><li>Initial velocity ,u = 40m/s</li><li>Final velocity ,v = 0m/s (as the stone reached its maximum height)</li><li>Acceleration due to gravity ,g = 10m/s²</li></ul><p></p><p><strong><u>To </u></strong><strong><u><a href="https://interviewquestions.tuteehub.com/tag/find-11616" style="font-weight:bold;" target="_blank" title="Click to know more about FIND">FIND</a></u></strong><strong><u>:</u></strong><strong><u>-</u></strong></p><p></p><ul><li>Maximum height ,h </li><li>Net Displacement ,S</li><li>Total distance covered by ball , d</li></ul><p></p><p></p><p><strong><u>Solution</u></strong><strong><u>:</u></strong><strong><u>-</u></strong><strong><u> </u></strong></p><p></p><p>Firstly we calculate the maximum height attained by the stone when it is thrown . Using 3rd equation of motion </p><p></p><ul><li><strong>v²</strong><strong> </strong><strong>=</strong><strong> </strong><strong>u²</strong><strong> </strong><strong>+</strong><strong> </strong><strong>2</strong><strong>g</strong><strong>h</strong></li></ul><p>where,</p><ul><li>v denote final velocity</li><li>u denote initial velocity</li><li>g denote acceleration due to gravity</li><li>h denote maximum height attained by stone</li></ul><p>Substitute the value we get</p><p></p><p>→ 0² = 40² + 2 (-10) × h</p><p></p><p>→ 0 = 1600 -20h</p><p></p><p>→ -1600 = -20h</p><p></p><p>→ 20h = 1600</p><p></p><p>→ h = 1600/20 </p><p></p><p>→ h = 80 m</p><p></p><ul><li><strong><u>Hence</u></strong><strong><u>,</u></strong><strong><u> </u></strong><strong><u>the </u></strong><strong><u>maximum</u></strong><strong><u> </u></strong><strong><u>height</u></strong><strong><u> attained</u></strong><strong><u> </u></strong><strong><u>by </u></strong><strong><u>the </u></strong><strong><u>stone</u></strong><strong><u> </u></strong><strong><u>is </u></strong><strong><u>8</u></strong><strong><u>0</u></strong><strong><u> </u></strong><strong><u>metres</u></strong></li></ul><p></p><p> The total distance covered by the stone when it falls back to the ground is <strong>8</strong><strong>0</strong><strong> </strong><strong>+</strong><strong> </strong><strong>8</strong><strong>0</strong><strong> </strong><strong>=</strong><strong> </strong><strong>1</strong><strong>6</strong><strong>0</strong><strong> </strong><strong>metres</strong><strong> </strong></p><p></p><p>And the net displacement of the stone is <strong>0</strong><strong> </strong><strong>metres</strong><strong> </strong>(as the stone reached its initial position).</p></body></html>

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⠀→ A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone when it falls back to the ground ?​

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⠀→ A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone when it falls back to the ground ?