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Projectile Motion

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New answer posted

6 days ago

Projectile Motion Class 11th

A particle of mass m is projected with speed v at an angle of 30° with the horizontal, find its angular momentum about point of projection when it reaches its maximum height.

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alok kumar singh

Contributor-Level 10

Velocity at maximum height = vcoss30°

L = m (vcos30) H

$= m v (\frac{\sqrt[]{3}}{2}) * \frac{v^{2} s i n^{2} 30}{2 g}$

$= \sqrt[]{3} \frac{m v^{3}}{16 g}$

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New answer posted

2 weeks ago

Projectile Motion Class 11th

The trajectory of a projectile in an vertical plane is y = ax - bx², where a and b are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection q and the maximum height attained H are respectively given by:

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

$y = α x - β x^{2}$

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

$\Rightarrow x = \frac{α}{2 β}$

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

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

2 weeks ago

Projectile Motion Class 11th

A fighter jet is flying horizontally at a certain with a speed of 200ms^-1. When it passes directly overhead ad anti-aircraft gun, a bullet is fired from the gun, at an angle (θ)with the horizontal, to hit the jet. If the bullet speed is 400m/s, the value of (θ)will be _____________.

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Raj Pandey

Contributor-Level 9

B → fighter jet

A → anti-Air craft gun

Draw velocity diagram of A w.r.t. B

If A hits B

Then Relative velocity perpendicular to the line joining A to B will be zero.

That means 400 cos q = 200

&nbs

...more

B → fighter jet

A → anti-Air craft gun

Draw velocity diagram of A w.r.t. B

If A hits B

Then Relative velocity perpendicular to the line joining A to B will be zero.

That means 400 cos q = 200

$\begin{array}{l} c o s θ = \frac{1}{2} \\ θ = 60 ° \end{array}$

less

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New answer posted

2 weeks ago

Projectile Motion Class 11th

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

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

According to question, we can write

$R = \frac{u^{2} s i n 2 θ}{g} \Rightarrow u^{2} = g R_{m a x}, w h e r e θ = 45 °$

$\Rightarrow H_{m a x} = \frac{u^{2}}{2 g} = \frac{g R_{m a x}}{2 g} = \frac{R_{m a x}}{2} = \frac{100}{2} = 50 m$

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New answer posted

2 weeks ago

Projectile Motion Class 11th

A projectile travels at θ below horizontal 1 sec after projection as shown. What is the value of θ?

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Vishal Baghel

Contributor-Level 10

$t a n θ = \frac{v_{y}}{v_{x}} = \frac{u_{y} - g t}{u_{x}}$

$\Rightarrow t a n θ = | \frac{8 - 10}{4} | = | - \frac{1}{2} |$

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New answer posted

3 weeks ago

Projectile Motion Class 11th

A grass hopper can jump maximum distance 1.6 m. It spends negligible time on ground. How far can it go in 2√2s (in meters)?

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

Maximum range is obtained at 45?
u²/g = 1.6 or u = 4m/s
T = (2u sin 45? )/g = (2*4* (1/√2)/10 = 0.4√2s
Number of jumps in a given time, n = t/T = 2√2 / 0.4√2 = 5

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New answer posted

3 weeks ago

Projectile Motion Class 11th

A bullet is fired from a gun at the speed of 280 m/s in the direction $30^{?}$ above the horizontal. The maximum height attained by the bullet is $(g = 9.8 {m s}^{- 2}, s i n ? 30^{?} = 0.5)$

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Raj Pandey

Contributor-Level 9

$h_{max} = \frac{u^{2} {s i n}^{2} ? θ}{2 g} = \frac{280 * 280}{2 * 9.8} * \frac{1}{4}$

=1000 m

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New answer posted

4 weeks ago

Projectile Motion Class 11th

Two transparent media A and B are separated by a plane boundary. The speed of light in those media are $1.5 * 10^{8} m / s$ and $2.0 * 10^{8} m / s$ , respectively. The critical angle for a ray of light for these two media is

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Raj Pandey

Contributor-Level 9

$i_{c} = {s i n}^{- 1} ? (\frac{3}{4})$

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New answer posted

4 weeks ago

Projectile Motion Class 11th

A bullet is fired from a gun at the speed of $280 m / s$ in the direction $30^{\circ}$ above the horizontal. The maximum height attained by the bullet is

$(g = 9.8 {m s}^{- 2}, s i n ? 30^{?} = 0.5)$

0 Follower 11 Views

alok kumar singh

Contributor-Level 10

Potential energy stored in the spring $= \frac{1}{2} {k x}^{2}$

$\begin{array}{r} Now \frac{1}{2} k (2)^{2} = U \\ & \frac{1}{2} k (8)^{2} = U^{'} (say) \\ \Rightarrow U^{'} = \frac{64}{4} U = 16 U \end{array}$

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New answer posted

a month 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:

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

$y = α x - β x^{2}$

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

$\Rightarrow x = \frac{α}{2 β}$

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

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

0 0

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