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Byju's Answer

Standard XII

Physics

The 'e' Equation

Jai is standi...

Question

Jai is standing of the top of building of height $25 m$ he want to throw his gun to Veeru who stands on top of another building of height $20 m$ at distance $15 m$ from first building. For which horizontal speed of projection, it is possible?

5 m s^{- 1}

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10 m s^{- 1}

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15 \sqrt{2} m s^{- 1}

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20 m s^{- 1}

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Solution

The correct option is C $15 \sqrt{2} m s^{- 1}$

$u_{x} = ?$
$u_{y} = 0$
$S_{y} = 5$ & $S_{x} = 15$
$S_{y} = u_{y} t + a_{y} t^{2}$
$\Rightarrow 5 = g \times t^{2} \Rightarrow t = \sqrt{\frac{5}{g}}$
$S_{x} = u_{x} t + a_{x} t^{2}$
$\Rightarrow 15 = u_{x} \sqrt{\frac{5}{g}} + O$
$\Rightarrow u_{x} = 15 \sqrt{\frac{10}{5}} = 15 \sqrt{2}$ m/s.

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Jai is standing of the top of building of height 25 m he want to throw his gun to Veeru who stands on top of another building of height 20 m at distance 15 m from first building. For which horizontal speed of projection, it is possible?

The correct option is C 15 √2ms−1ux=?uy=0Sy=5 & Sx=15Sy=uyt+ayt2⇒5=g×t2⇒t=√5gSx=uxt+axt2⇒15=ux√5g+O⇒ux=15√105=15√2m/s.

Jai is standing of the top of building of height $25 m$ he want to throw his gun to Veeru who stands on top of another building of height $20 m$ at distance $15 m$ from first building. For which horizontal speed of projection, it is possible?