Projectile Motion

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New answer posted•a year ago

Projectile Motion Class 11th

The trajectory of a projectile in an vertical plane is y = αx - βx², where α and β are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection θ and the maximum height attained H are respectively given by:

18 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$y = α x - β x^{2}$

$\Rightarrow \frac{d y}{d x} = α - 2 β x = 0$

$\Rightarrow x = \frac{α}{2 β}$

$y_{m a x} = α \times \frac{α}{2 β} - β \times {(\frac{α}{2 β})}^{2} = \frac{α^{2}}{4 β}$

Range = $2 x = \frac{α}{β} = \frac{2 u^{2} s i n θ . c o s θ}{g}$

On comparing with

y = x tan θ - $\frac{g x^{2}}{2 u^{2} c o s^{2} θ}$

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New answer posted•a year ago

Projectile Motion Class 11th

If the initial velocity in horizontal direction of a projectile is unit vector $\hat{i}$  and the equation of trajectory is y = 5x(1 – x). The y component vector of the initial velocity is ________ $\hat{j}$ .

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

y = x5 (1 – x) = x tan θ $(1 - \frac{x}{R})$

tan = 5, R = 1

$s i n θ = \frac{5}{\sqrt[]{26}}, c o s θ = \frac{1}{\sqrt[]{26}}$

$R = \frac{u^{2} s i n 2 θ}{g} = 1$

$\Rightarrow u^{2} = 26 \Rightarrow u = \sqrt[]{26} m / s$

y – component of initial velocity

= u sin θ

= $= \sqrt[]{26} \times \frac{5}{\sqrt[]{26}}$

= 5 m/s

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New answer posted•a year ago

Projectile Motion Class 11th

Two projectiles thrown at $30 ° a n d 45 °$  with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

H₁ = H₂

$\frac{u_{1}^{2} s i n^{2} θ_{1}}{2 g} = \frac{u_{2}^{2} s i n^{2} θ_{2}}{2 g}$

$\Rightarrow u_{1}^{2} {(s i n 30 °)}^{2} = {u_{2}}^{2} {(s i n 45 °)}^{2}$

$\Rightarrow$ $\frac{u_{1}}{u_{2}}$ $=2$

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New answer posted•a year ago

Projectile Motion Class 11th

A monkey of mass 50kg climbs on a rope which can withstand the tension (T) of 350 N. If monkey initially climbs down with an acceleration of 4 m/s² and then climbs up with an acceleration of 5 m/s². Choose the correct option (g = 10 m/s²).

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

For climbing downward

50 g – T = 50a

T = 500 – 50 * 4

= 300 N < 350 N

for climbing upwards

T – 50 g = 50 a

T – 500 = 50 * 5

T = 750 N > 350 N

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New answer posted•a year ago

Projectile Motion Class 11th

A helicopter is flying horizontally with a speed ‘v’ at an altitude ‘h’ has to drop a food packet for a man on the ground. What is the distance of helicopter from the man when the food packet is dropped?

4 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

$h = \frac{1}{2} g t^{2}$

or $x = v \sqrt[]{\frac{2 h}{g}}$

So, x = vt

or, $x = v \frac{2 h}{g}$

So, dist of man from helicopter is

$\sqrt[]{h^{2} + x^{2}} = \sqrt[]{h^{2} + v^{2} \frac{2 h}{g}} = \sqrt[]{\frac{2 v^{2} h}{g} + h^{2}}$

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New answer posted•a year ago

Projectile Motion Class 11th

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

Reason R : Product of said heights.

$h_{1} h_{2} = (\frac{u^{2} s i n^{2} θ}{2 g}) . (\frac{u^{2} c o s^{2} θ}{2 g})$

Choose the correct answer :

6 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

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

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

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

$h_{1} h_{2} = \frac{1}{4^{2}} R^{2}$

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

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

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New answer posted•a year ago

Projectile Motion Motion in a Plane

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)

14 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

d = R

$60 = R [\frac{3 π}{4}]$

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

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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New answer posted•a year ago

Projectile Motion Class 11th

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

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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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New answer posted•a year ago

Projectile Motion Class 11th

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 _______.

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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

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New answer posted•a year ago

Projectile Motion Class 11th

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)

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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°

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

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

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