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

Potential energy stored in the spring = 1 2 k x 2 Now 1 2 k ( 2 ) 2 = U & 1 2 k ( 8 ) 2 = U ' (say) ⇒ U ' = 64 4 U = 16 U

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

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

Option 1 - <math> <mn>3000</mn> <mtext> </mtext> <mi mathvariant="normal">m</mi> </math>

Option 2 - <math> <mn>2800</mn> <mtext> </mtext> <mi mathvariant="normal">m</mi> </math>

Option 3 - <math> <mn>2000</mn> <mtext> </mtext> <mi mathvariant="normal">m</mi> </math>

Option 4 - <math> <mn>1000</mn> <mtext> </mtext> <mi mathvariant="normal">m</mi> </math>

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Answered by

Alok Kumar Singh | Contributor-Level 10

5 months ago

Correct Option - 1

Detailed Solution:

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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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

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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