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Physics Motion in Straight Line

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3 months ago

Physics Motion in Straight Line Class 11th

A juggler throws ball vertically upwards with same initial velocity in air. When the first ball reaches its highest positions he throws the next ball. Assuming the juggler throws n balls per second, the maximum height the balls can reach is

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alok kumar singh

Contributor-Level 10

Time to reach at max height t = $\frac{u}{g} n o .$ of balls thrown in 1 sec = n.

So, time taken by each ball to reach max^m height, = $\frac{1}{n} s e c$

i.e. $\frac{u}{g} = \frac{1}{n} \Rightarrow u = \frac{g}{n}$ So, h_max = $\frac{u^{2}}{2 g}$

= $\frac{g^{2}}{2 g n^{2}}$

$= \frac{g}{2 n^{2}}$

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3 months ago

Physics Motion in Straight Line Class 11th

A particle is moving in a straight line such that its velocity is increasing at 5 ms^-1 per meter. The acceleration of the particle is _________ms^-2 at a point where its velocity is 20 ms^-1.

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Payal Gupta

Contributor-Level 10

$t a n θ = \frac{d v}{d x} = 5 (m / s) / m$

$\Rightarrow d v = 5 d x \Rightarrow \frac{d v}{d t} = 5 \frac{d x}{d t}$

$\Rightarrow a = 5 v = 5 * 20 = 100 m / s^{2}$

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Physics Motion in Straight Line Class 11th

A force on an object of mass 100g is $(10 \hat{i} + 5 \hat{j}) N .$ The position of that object at t = 2s is $(a \hat{i} + b \hat{j}) m$ after starting from rest. The value of $\frac{a}{b}$ will be

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Payal Gupta

Contributor-Level 10

$\vec{F} = 10 \hat{i} + 5 \hat{j}$

$\vec{a} = \frac{10}{m} \hat{i} + \frac{5}{m} \hat{j}$

$= \frac{10}{0.1 k g} \hat{i} + \frac{5}{0.1} \hat{j}$

$\vec{a} = 100 \hat{i} + 50 \hat{j}$

$a_{x} = 100, a_{y} = 50$

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

$= 0 * 2 + \frac{1}{2} * 100 * 2 * 2$

Sx = 200 m

$\frac{a}{b} = \frac{200}{100} = 2$

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3 months ago

Physics Motion in Straight Line Class 11th

An object of mass 5kg is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of 10 N throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use g = 10ms^-2].

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Payal Gupta

Contributor-Level 10

$a = \frac{m g + 10}{g} = \frac{5 * 10 + 10}{10}$

$\Rightarrow a = 6 m / s e c^{2}$

v = u + at

0 = u – 6t

$t_{A} = \frac{u}{6}$ (time of Ascent)

$= u * \frac{u}{6} - \frac{u}{2} * 6 * \frac{u}{6} * \frac{u}{6}$

= $\frac{u^{2}}{6} - \frac{u^{2}}{12}$

$h = \frac{u^{2}}{12}$

for time of descent

$a = \frac{50 - 10}{10}$

a = 4 m /sec²

$\Rightarrow \frac{u^{2}}{12} = 0 * t + \frac{1}{2} * 4 t^{2}$

$t_{B} = \frac{u}{\sqrt[]{24}}$

$\frac{t_{A}}{t_{B}} = \frac{u * 24}{6 * u} =$ $\sqrt[]{2}$

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