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Physics NCERT Exemplar Solutions Class 11th Chapter Five Physics Class 12th Motion in a Straight Line Physics NCERT Exemplar Solutions Class 12th Chapter Three

Two balls A and B placed at the top of 180m tall tower. Ball A is from the top at t = 0s. Ball B is thrown vertically down with an initial velocity ‘u’ at t = 2s. After a certain time, both balls meet 100m above the ground. Find the value of ‘u’ in ms^-¹.

[use g = 10 ms^-²] :

Option 1 -

Option 2 -

Option 3 -

Option 4 -

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

Let 't' be the time taken by B to meat A.

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec²

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

80 = 0 * t + $\frac{1}{2} (10) {(2 + t)}^{2}$

16 = (2 + t)²

t + 2 = 4

t = 2sec

Now for B

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

...more

Let 't' be the time taken by B to meat A.

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec²

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

80 = 0 * t + $\frac{1}{2} (10) {(2 + t)}^{2}$

16 = (2 + t)²

t + 2 = 4

t = 2sec

Now for B

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

$? 80 = u * 2 + \frac{1}{2} * 10 t^{2}$

80 = 2u + 5 * 2 * 2

80 = 2u + 20

2u = 60

u = 30m/sec

less

Let 't' be the time taken by B to meat A.                                                                for A, u = 0t = [2 + t]h = 80ma = 10 m/sec2h = ut + <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1814255368"> </o:OLEObject></xml><! [endif]->  <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> 80 = 0 * t + <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mrow> <mo> (</mo> <mrow> <mn>1</mn> <mn>0</mn> </mrow> <mo>)</mo> </mrow> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mn>2</mn> <mo>+</mo> <mi>t</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1814255369"> </o:OLEObject></xml><! [endif]->16 = (2 + t)2t + 2 = 4 t = 2secNow for Bh = ut + <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1814255370"> </o:OLEObject></xml><! [endif]->  <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>               <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1028" DrawAspect="Content" ObjectID="_1814255371"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mo>? </mo> <mn>8</mn> <mn>0</mn> <mo>=</mo> <mi>u</mi> <mo>*</mo> <mn>2</mn> <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>  80 = 2u + 5 * 2 * 2 80 = 2u + 20  2u = 60  u = 30m/sec

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Two balls A and B placed at the top of 180m tall tower. Ball A is from the top at t = 0s. Ball B is thrown vertically down with an initial velocity ‘u’ at t = 2s. After a certain time, both balls meet 100m above the ground. Find the value of ‘u’ in ms^-¹. [use g = 10 ms^-2]

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Let ‘t’ be the time taken by B to meat

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec²

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

80 = 0 × t + $\frac{1}{2} (10) {(2 + t)}^{2}$

16 = (2 + t)²

t + 2 = 4

t = 2sec

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

$\Rightarrow 80 = u \times 2 + \frac{1}{2} \times 10 t^{2}$

80 = 2u + 5 × 2 × 2

80 = 2u + 20

2u = 60

u = 30m/sec

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Two balls A and B placed at the top of 180m tall tower. Ball A is from the top at t = 0s. Ball B is thrown vertically down with an initial velocity ‘u’ at t = 2s. After a certain time, both balls meet 100m above the ground. Find the value of ‘u’ in ms^-¹.

[use g = 10 ms^-²] :

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Two balls A and B placed at the top of 180m tall tower. Ball A is from the top at t = 0s. Ball B is thrown vertically down with an initial velocity ‘u’ at t = 2s. After a certain time, both balls meet 100m above the ground. Find the value of ‘u’ in ms-1. [use g = 10 ms-2] :

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Your Question

Two balls A and B placed at the top of 180m tall tower. Ball A is from the top at t = 0s. Ball B is thrown vertically down with an initial velocity ‘u’ at t = 2s. After a certain time, both balls meet 100m above the ground. Find the value of ‘u’ in ms^-¹.

[use g = 10 ms^-²] :