From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in __________ s.

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Answered by

Vishal Baghel | Contributor-Level 10

3 months ago

For first ball

s = ut + $\frac{1}{2} a t^{2}$

$\Rightarrow - h = u \times 6 - \frac{1}{2} \times 10 \times 6 \times 6$

$\Rightarrow - h = 6 u - 180$

h = 180 - 6u -(1)

for second boy

$h = u t \frac{1}{2} a t^{2}$

$h = u \times 1.5 + \frac{1}{2} \times 10 \times 1.5 \times 1.5$

h = 1.5u + 11.25 -(2)

from (1) & (2)

180 - 6u = 1.5u + 11.25

7.5u = 180 - 11.25

$u = \frac{168.75}{7.5} = 22.5$

h = 180 - 6u

= 180 - 6 × 22.5

45m

for third ball

$h = u t + \frac{1}{2} a t^{2}$

$h = 0 \times t + \frac{1}{2} \times 10 t^{2}$

$\Rightarrow 45 = 5 t^{2}$

t² = 9

t = 3 sec

For first ball s = ut +   <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>a</mi> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math> <math> <mrow> <mo>⇒</mo> <mo>−</mo> <mi>h</mi> <mo>=</mo> <mi>u</mi> <mo>×</mo> <mn>6</mn> <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>6</mn> <mo>×</mo> <mn>6</mn> </mrow> </math> <math> <mrow> <mo>⇒</mo> <mo>−</mo> <mi>h</mi> <mo>=</mo> <mn>6</mn> <mi>u</mi> <mo>−</mo> <mn>1</mn> <mn>8</mn> <mn>0</mn> </mrow> </math>      h = 180 - 6u                     -(1)for second boy <math> <mrow> <mi>h</mi> <mo>=</mo> <mi>u</mi> <mi>t</mi> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>a</mi> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math> <math> <mrow> <mi>h</mi> <mo>=</mo> <mi>u</mi> <mo>×</mo> <mn>1</mn> <mo>.</mo> <mn>5</mn> <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>1</mn> <mo>.</mo> <mn>5</mn> <mo>×</mo> <mn>1</mn> <mo>.</mo> <mn>5</mn> </mrow> </math>             h = 1.5u + 11.25               -(2)from (1) & (2)180 - 6u = 1.5u + 11.257.5u = 180 - 11.25 <math> <mrow> <mi>u</mi> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>6</mn> <mn>8</mn> <mo>.</mo> <mn>7</mn> <mn>5</mn> </mrow> <mrow> <mn>7</mn> <mo>.</mo> <mn>5</mn> </mrow> </mfrac> <mo>=</mo> <mn>2</mn> <mn>2</mn> <mo>.</mo> <mn>5</mn> </mrow> </math>    h = 180 - 6u<img >= 180 - 6 × 22.545mfor third ball <math> <mrow> <mi>h</mi> <mo>=</mo> <mi>u</mi> <mi>t</mi> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>a</mi> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math> <math> <mrow> <mi>h</mi> <mo>=</mo> <mn>0</mn> <mo>×</mo> <mi>t</mi> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>×</mo> <mn>1</mn> <mn>0</mn> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>      <math> <mrow> <mo>⇒</mo> <mn>4</mn> <mn>5</mn> <mo>=</mo> <mn>5</mn> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>          t2 = 9t = 3 sec

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