A stone is dropped from the top of a building. When it crosses a point 5m below the top, another stone starts to fall from a point 25m below the top. Both stones reach the bottom of building simultaneously. The height of the building is

Correct Option - 2


Detailed Solution:

When second stone is released

Equation for first ball:

$x+20=10t+\frac{1}{2}g{t}^2$

Equation for second ball :

$x=\frac{1}{2}g{t}^2$

Using these two equation

20 = 10t t = 2 sec

$x=\frac{1}{2}\times 10\times 4=20m$

H = 20 + 20 + 5 = 45 m.

<p>When second stone is released</p><p>Equation for first ball:</p><p><math><mrow><mi>x</mi><mo>+</mo><mn>2</mn><mn>0</mn><mo>=</mo><mn>1</mn><mn>0</mn><mi>t</mi><mo>+</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mi>g</mi><msup><mrow><mi>t</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></math></p><p>Equation for second ball :</p><p><math><mrow><mi>x</mi><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mi>g</mi><msup><mrow><mi>t</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></math></p><p>Using these two equation</p><p>20 = 10t t = 2 sec</p><p><math><mrow><mi>x</mi><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mo>×</mo><mn>1</mn><mn>0</mn><mo>×</mo><mn>4</mn><mo>=</mo><mn>2</mn><mn>0</mn><mi>m</mi></mrow></math></p><p>H = 20 + 20 + 5 = 45 m.</p><div class="photo-widget-full"><div class="figure"><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1755056681phpsX5pXy.jpeg" alt="" width="206" height="156" loading="lazy"></picture></div></div>

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