A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height $\frac{h}{3}$ while going up and coming down respectively.

Option 1 - <math> <mrow> <mfrac> <mrow> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> <mo>−</mo> <mn>1</mn> </mrow> <mrow> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> <mo>+</mo> <mn>1</mn> </mrow> </mfrac> </mrow> </math>

Option 2 - <math> <mrow> <mfrac> <mrow> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> <mo>−</mo> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> <mrow> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> <mo>+</mo> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> </mrow> </math>

Option 3 - <math> <mrow> <mfrac> <mrow> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> <mo>−</mo> <mn>1</mn> </mrow> <mrow> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> <mo>+</mo> <mn>1</mn> </mrow> </mfrac> </mrow> </math>

Option 4 - <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math>

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Vishal Baghel | Contributor-Level 10

6 months ago

Correct Option - 2

Detailed Solution:

$h = \frac{u^{2}}{2 g} \Rightarrow u = \sqrt[]{2 g h}$

$\frac{h}{3} = \sqrt[]{2 g h} t - 5 t^{2}$

$5 t^{2} - \sqrt[]{20 h} t + \frac{h}{3} = 0$

$t = \frac{\sqrt[]{20 h} \pm \sqrt[]{20 h - 4 * 5 * \frac{h}{3}}}{10}$

$∴ \frac{t_{1}}{t_{2}} = \frac{\sqrt[]{20 h} - \sqrt[]{\frac{40}{3} h}}{\sqrt[]{20 h} + \sqrt[]{\frac{40}{3} h}} = \frac{1 - \sqrt[]{\frac{2}{3}}}{1 + \sqrt[]{\frac{2}{3}}} = \frac{\sqrt[]{3} - \sqrt[]{2}}{\sqrt[]{3} + \sqrt[]{2}}$

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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] :

Alok Kumar Singh

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

Payal Gupta

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

Now for B

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

A scooter accelerates from rest for time t₁ at constant rate a₁ and then retards at constant rate a₂ for time t₂ and comes to rest. The correct value of $\frac{t_{1}}{t_{2}}$ will be

Payal Gupta

$t a n α = \frac{v_{m a x}}{t_{1}} = a_{1}$

$t a n β = \frac{v_{m a x}}{t_{2}} = a_{2}$

$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{t_{2}}{t_{1}} \Rightarrow \frac{t_{1}}{t_{2}} = \frac{a_{2}}{a_{1}}$

A Tennis ball is released from a height $h$ and after freely falling on a wooden floor it rebounds and reaches height $\frac{h}{2}$ . The velocity versus height of the ball during its motion may be represented graphically by:

(graphs are drawn schematically and on not to scale)

Alok Kumar Singh

$\begin{array}{r} \Rightarrow V^{2} = U^{2} + 2 g S \\ \Rightarrow V^{2} = 0 + 2 g (h - y) \\ \Rightarrow V^{2} = 2 g h - 2 g y \\ \Rightarrow V = \sqrt[]{2 g h - 2 g y} \end{array}$

A ball is thrown vertically upwards with a velocity of 19.6 ms^-1 from the top of a tower. The ball strikes the ground after 6s. The height from the ground up to which the ball can rise will be $(\frac{k}{5}) m .$ The value of k is_______(use g = 9.8m/s²)

Vishal Baghel

h = 39.2 + 19.6 = 58.8 m

Height above tower

$h' = \frac{u^{2}}{2 g} = \frac{19.6 \times 19.6}{2 \times 9.8}$

= 19.6

$A s, \frac{k}{5} = 78.4 \Rightarrow k = 392$

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A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height h3 while going up and coming down respectively.