A fighter plane moving with a speed of $50 \sqrt{2} m / s$ is at an elevation of 1000m above the ground. It releases a bomb upward at angle of $45^{0}$ with the vertical.
Find maximum height of the bomb above ground.

1025 m

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1125 m

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1075 m

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1175 m

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Solution

The correct option is B $1125 m$
$v = 50 \sqrt{2} m / s$
$n = 1000 m$

max. height abv aeroplane $= \frac{{(50 \sqrt{2} \times \frac{1}{\sqrt{2}})}^{2}}{2 \times 10}$

$= \frac{50 \times 50}{20}$

$= 125 m$

$∴$ total max. h $1000 + 125 = 1125 m$

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A fighter plane moving with a speed of 50√2m/s is at an elevation of 1000m above the ground. It releases a bomb upward at angle of 450 with the vertical.Find maximum height of the bomb above ground.

The correct option is B 1125mv=50√2m/s n=1000mmax. height abv aeroplane =(50√2×1√2)22×10 =50×5020 =125m ∴ total max. h 1000+125=1125m

A fighter plane moving with a speed of $50 \sqrt{2} m / s$ is at an elevation of 1000m above the ground. It releases a bomb upward at angle of $45^{0}$ with the vertical.
Find maximum height of the bomb above ground.

The correct option is B $1125 m$
$v = 50 \sqrt{2} m / s$
$n = 1000 m$

max. height abv aeroplane $= \frac{{(50 \sqrt{2} \times \frac{1}{\sqrt{2}})}^{2}}{2 \times 10}$

$= \frac{50 \times 50}{20}$

$= 125 m$

$∴$ total max. h $1000 + 125 = 1125 m$