A bullet is fired from a gun at the speed of 280 m/s in the direction <math> <msup> <mrow> <mrow> <mn>30</mn> </mrow> </mrow> <mrow> <mrow> <mo>?</mo> </mrow> </mrow> </msup> </math> above the horizontal. The maximum height attained by the bullet is <math> <mfenced separators="|"> <mrow> <mrow> <mi mathvariant="normal">g</mi> <mo>=</mo> <mn>9.8</mn> <msup> <mrow> <mrow> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">s</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> <mo>,</mo> <mi mathvariant="normal">s</mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">n</mi> <mo>?</mo> <msup> <mrow> <mrow> <mn>30</mn> </mrow> </mrow> <mrow> <mrow> <mo>?</mo> </mrow> </mrow> </msup> <mo>=</mo> <mn>0.5</mn> </mrow> </mrow> </mfenced> </math>

h max = u 2 s i n 2 ? θ 2 g = 280 × 280 2 × 9.8 × 1 4 =1000 m

A bullet is fired from a gun at the speed of 280 m/s in the direction <math> <msup> <mrow> <mrow> <mn>30</mn> </mrow> </mrow> <mrow> <mrow> <mo>?</mo> </mrow> </mrow> </msup> </math> above the horizontal. The maximum height attained by the bullet is <math> <mfenced separators="|"> <mrow> <mrow> <mi mathvariant="normal">g</mi> <mo>=</mo> <mn>9.8</mn> <msup> <mrow> <mrow> <mtext> </mtext> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">s</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> <mo>,</mo> <mi mathvariant="normal">s</mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">n</mi> <mo>?</mo> <msup> <mrow> <mrow> <mn>30</mn> </mrow> </mrow> <mrow> <mrow> <mo>?</mo> </mrow> </mrow> </msup> <mo>=</mo> <mn>0.5</mn> </mrow> </mrow> </mfenced> </math>

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Projectile Motion Physics Motion in Plane Physics 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 $(g = 9.8 {m s}^{- 2}, s i n ? 30^{?} = 0.5)$

Option 1 -

$3000 m$

Option 2 -

$2800 m$

Option 3 -

$2000 m$

Option 4 -

$1000 m$

2 Views | Posted 3 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

3 weeks ago

Correct Option - 4

Detailed Solution:

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

=1000 m

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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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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.

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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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According to question, we can write

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

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