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

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

Question Image
Grade:9

1 Answers

Eshan
askIITians Faculty 2095 Points
2 years ago
Dear student,

Since the horizontal range is given by-

R=\dfrac{2u^2sin\theta cos\theta}{g},
we have360=\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

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