Physics Motion in Plane

Given below are two statements. One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A: Two identical balls A and B thrown with same velocity 'u' at two different angles with horizontal attained the same range R. If A and B reached the maximum height h1 and h2 respectively, then

Reason R : Product of said heights.

Choose the correct answer :

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alok kumar singh

Contributor-Level 10

First angle, θ₁= θ, $R = \frac{u^{2} s i n 2 θ}{g}$

Another angle (θ₂ = 90 - θ) for which range will be

Same as that of θ₁ = θ

$R^{1} = \frac{u^{2} s i n 2 (90 - θ)}{g} = \frac{u^{2}}{g} s i n 2 θ = R$

at $θ_{1} = θ, h_{1} = \frac{u^{2} s i n^{2} θ}{2 g}$

$& θ_{2} = 90 - θ, h_{2} = \frac{u^{2} s i n^{2} θ_{2}}{2 g} = \frac{u^{2} c o s^{2} θ}{2 g}$

$\Rightarrow h_{1} h_{2} = \frac{1}{4^{2}} {[\frac{u^{2} s i n 2 θ}{g}]}^{2}$

$R = 4 \sqrt[]{h_{1} h_{2}}$

So, Both statement is true & Reason is correct explanation for statement 1.

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A projectile is launched at an angle 'α' with the horizontal with a velocity 20 ms^-¹. After 10 s, it inclination with horizontal is β. The value of tanβ will be: (g = 10ms^-²).

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Contributor-Level 10

We know that horizontal speed will remain constant

20 cos = v cos ……… (i)

Along y-axis

$U_{y} = 20 s i n α, V_{y} = v s i n β$

$V_{y} = V_{y} + a_{y} t$

$V s i n β = 20 s i n α - 10 * 10$ ……. (ii)

$\Rightarrow (i i) \div (i)$

$\Rightarrow t a n β = t a n α - 5 s e c α$

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A person moved from A to B on a circular path as shown in figure. If the distance travelled by him is 60m, then the magnitude of displacement would be:

(Given $c o s 135 °$ = -0.7)

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Contributor-Level 10

d = R

$60 = R [\frac{3 π}{4}]$

$R = \frac{60 * 4}{3 π} = \frac{80}{π} m$

Displacement = $\sqrt[]{R^{2} + R^{2} - 2 R^{2} c o s 135}$ $^{}^{}^{}$

= $\sqrt[]{2 R^{2} - 2 R^{2} (- 0.7)}$ $^{}^{}$

= $\sqrt[]{3.4 R^{2}}$ $^{}$

= $\sqrt[]{3.4 {(\frac{80}{4})}^{2}}$ ${\frac{}{}}^{}$

47 m

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A ball is projected from the ground with a speed 15 ms^-1 at an angle θ with horizontal so that its range and maximum height are equal. Then 'tanθ' will be equal to

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Contributor-Level 10

$H = \frac{u^{2} s i n^{2} θ}{2 g} a n d R = \frac{u^{2} s i n 2 θ}{g}$

According to question

$R = H \Rightarrow \frac{u^{2} s i n^{2} θ}{2 g} = \frac{2 u^{2} s n θ c o s θ}{g} \Rightarrow t a n θ = 4$

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The sum of two forces $\vec{P}$ and $\vec{Q}$ is $\vec{R}$ such that $| \vec{R} | = | \vec{P} |$ . The angle $θ$ (in degrees) that the resultant of $2 \vec{P}$ and $\vec{Q}$ will make with $\vec{Q}$ is,

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alok kumar singh

Contributor-Level 10

$| \overset{?}{P} + \overset{?}{Q} | = | \overset{?}{P} |$

^{$P^{2} + Q^{2} + 2 P Q c o s ? θ = P^{2}$}

$\Rightarrow Q + 2 P c o s ? θ = 0$

$\Rightarrow c o s ? θ = - \frac{Q}{2 P}$

$t a n ? α = \frac{2 P s i n ? θ}{2 P c o s ? θ + Q} = \infty$

$? [2 P c o s ? θ + Q = 0]$

α = 90^{?}

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Two inclined planes are placed as shown in figure. A block in Projected from the Point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top Point B at a height 10m. After reaching the Point B the block slides down on inclined plane BC. Time it takes to reach to the point C from point A is t $t (\sqrt[]{2} + 1) s$ . The value of t is__________.

(use g = 10 m/s²)

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Contributor-Level 10

From conservation of Energy

$\frac{1}{2} m v_{0}^{2} = m g h$

$v_{0} = 10 \sqrt[]{2}$

For A to B

$a = - g s i n 45 ° = \frac{- 10}{\sqrt[]{2}}$

T = t₁ + t₂

$\begin{array}{l} = 2 \sqrt[]{2} + 2 \\ = 2 (\sqrt[]{2} + 1) \end{array}$

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A ball of mass m is thrown vertically upwards. Another ball of mass 2 is thrown at an angle q with vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is $\frac{1}{x} .$ The value of x is _______.

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Contributor-Level 10

$T_{1} = \frac{2 u}{g} T_{2} = \frac{2 V s i n θ}{g}$

$A s, T_{1} = T_{2} \Rightarrow \frac{2 u}{g} = \frac{2 V s i n θ}{g} \Rightarrow u = v s i n θ$ - (i)

$\frac{H_{1}}{H_{2}} = \frac{u^{2}}{2 g} * \frac{2 g}{v^{2} s i n^{2} θ} = {(\frac{u}{v s i n θ})}^{2} = 1$

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2 months ago

A body is projected from the ground at an angle of $45 °$ with horizontal. Its velocity after 2s is 20 ms^-1. The maximum height reached by the body during its motion is ______m. (use g = 10ms^-2)

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Contributor-Level 10

after 2 sec, v = 20 m/sec

$\begin{array}{l} \vec{v} = u c o s 45 ° \hat{i} + (u s i n 45 ° - g t) \hat{j} \\ v = 20 = \sqrt[]{{(u c o s 45 °)}^{2} + {(u s i n 45 ° - 20)}^{2}} \end{array}$

$\Rightarrow 400 = u^{2} + 400 - 40 u s i n 45 °$

u = 40 sin 45°

$H_{m a x} = \frac{u^{2} s i n^{2} 45}{2 g} = \frac{4 0^{2} s i n^{2} 45 ° * s i n^{2} 45}{2 g}$

$= \frac{40 * 40}{2 * 10} * \frac{1}{2} * \frac{1}{2} = 20 m$

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A projectile is projected with velocity of 25m/s at an angle θ with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of θ will be : &nb

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Contributor-Level 10

T = 2t = $\frac{}{} \frac{}{}$ $\frac{2 u s i n θ}{g} \Rightarrow t = \frac{u s i n θ}{g}$

R = u cos θ (T)

$\Rightarrow R = u c o s θ (\frac{2 u s i n θ}{g})$

$\Rightarrow R = u c o s θ 2 t$

$\Rightarrow c o s θ = \frac{R}{50 t} * \frac{t}{t}$

$\Rightarrow c o s θ = \frac{R t}{50 t^{2}}$

$\Rightarrow c o s θ = \frac{R}{50 t^{2}} * \frac{u s i n θ}{g}$

$\Rightarrow c o t θ = \frac{R}{50 t^{2}} * \frac{25}{10}$

cot θ = $\frac{R}{20 t^{2}}$ $θ = c o t^{- 1} (\frac{R}{20 t^{2}})$ $\frac{}{^{}}$

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2 months ago

A person can throw a ball upto a maximum range of 100m. How high above the ground he can the same ball?

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