Question

Chapter : Kinematics

Answer this with a Explanation :)

#No spams

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Answer 1

$\huge\sf\pink{Answer}$

☞ Your answer is 4×10³ m/s

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$\huge\sf\blue{Given}$

✭ Angle made while the missile is 45°

✭ It has to hit a target on the ground at 1600 km (Range)

✭ Gravitational Acceleration = 10 m/s²

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$\huge\sf\gray{To \:Find}$

◈ Velocity with which the missile has to be launched?

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$\huge\sf\purple{Steps}$

So here we shall first convet Range from Km to m

➝ 1 km = 1000 m

➝ 1600 km = 1000 × 1600

➝ 1600000 m

➝ 16 × 10⁵

We know that,

$\underline{\boxed{\sf Range = \dfrac{u^2 \ sin \ 2\theta}{g}}}$

On Substituting the given values,

➳ $\sf 16\times 10^5 = \dfrac{u^2 \ sin \ 2(45)}{10}$

➳ $\sf 16 \times 10^5 \times 10 = u^2 \ sin \ 90$

$\bigg\lgroup \sf sin \ 90^{\circ} = 1 \bigg\rgroup$

➳ $\sf 16\times 10^6 = u^2 \times 1$

➳ $\sf \sqrt{16\times 10^6} = u$

➳ $\sf \orange{u = 4\times 10^3 \ m/s}$

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Answer 2

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$\bigstar \mathtt{\huge\green{Question}}$

A missile is launched from the ground making an angle of 45° with the horizontal if it is required to hit the target on the ground at 1600 kilometres with what velocity should it be launched assume the acceleration due to gravity to be uniform throughout the motion of missile

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$\bigstar \mathtt{\huge\blue{given}}$

• angle of projection = 45 °
• Range of the projectile = 1600 km

we know that ,

1 km = 1000 m

1600 km = 1600000 m

1600 km = 16 × 10^5 m

Range of the projectile = 16× 10^5 m

• acceleration due to gravity = 10 m /s²

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$\bigstar \mathtt{\huge\pink{formula :}}$

Range of the projectile is given by the formula,

$\implies \boxed{ \mathtt{R = \dfrac{ {u}^{2} \sin(2 \theta) }{g} }}$

here ,

• R = range
• u = initial velocity
• theta = angle of projection
• g = acceleration due to gravity

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$\bigstar \mathtt{\huge\</u><u>o</u><u>r</u><u>a</u><u>n</u><u>g</u><u>e</u><u>{</u><u>S</u><u>olution :}}$

Range of the projectile is given by the formula,

$\implies \boxed{ \mathtt{R = \dfrac{ {u}^{2} \sin(2 \theta) }{g} }}$

we know that ,

$\implies \boxed{ \mathtt{sin2 \theta = 2sin \theta .cos \theta}}$

substituting this in the above equation we get ,

$\implies \mathtt{R = \dfrac{ {u}^{2}2 \sin( \theta) . \cos( \theta) }{g} }$

Substituting the given values in the above equation ,

$\implies \mathtt{R = \dfrac{ {u}^{2}2 \sin( 45) . \cos( 45) }{g} }$

$\implies \mathtt{R = \dfrac{ {u}^{2}2( \dfrac{1}{ \sqrt{2} } . \dfrac{1}{ \sqrt{2} } )}{g} }$

$\implies \mathtt{R = \dfrac{ {u}^{2} \cancel2( \dfrac{1}{ \cancel2} )}{g} }$

$\implies \mathtt{ {u}^{2} = 16 \times {10}^{5} \times10}$

$\implies \mathtt{u = \sqrt{16 \times {10}^{6} } }$

$\implies \mathtt{ \red{u = 4 \times {10}^{3} \frac{m}{s} }}$

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Hope it Helps : D