Grade 910 grade science

Q62............ class 10 physics..........................................................

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8 Years agoGrade 9

1 Answer

Eshan

8 Years ago

Dear student,

Since the horizontal range is given by-

$R=\dfrac{2u^2sin\theta cos\theta}{g}$ ,
we have $360=\dfrac{2u^2.\dfrac{1}{\sqrt{2}}.\dfrac{1}{\sqrt{2}}}{10}$
$\implies u=60m/s$
Vertical velocity component when fired from the trolley= $usin45^{\circ}=30\sqrt{2}m/s$

Horizontal velocity component when fired from trolley= $ucos 45^{\circ}+18=30\sqrt{2}m/s+18m/s$
Hence range of the bullet= $\dfrac{2u_xu_y}{g}=\dfrac{2\times 30\sqrt{2}(30\sqrt{2}+18)}{10}=512.73m$