Question

A fixed mortar fires a bomb at an angle of $$\displaystyle 53^{\circ}$$ above the horizontal with a muzzle velocity of $$80$$ $$\displaystyle ms^{-1}.$$ A tank is advancing directly towards the mortar on level ground at a constant speed of 5 m/s. The initial separation (at the instant mortar is fired ) between the mortar and tank, so that the tank would be hit is (Take $$\displaystyle \:g=10 \ ms^{-2}$$)

A
662.4 m
B
526.3 m
C
486.6 m
D
None of these

Solution

The correct option is D None of thesetime taken and range for projectile is $$t = \dfrac{2 u sin \theta}{g} = 16 sin 53^o$$$$R = \dfrac{u^2 sin 2 \theta}{g} = 640 sin 106^o$$ The initial separation  d should be equal to the range of mortar plus the distance traveled by tank. $$d = 5 \times 16 sin 53^o + 640 sin 106^o = 5 \times 0.8 + 640 \times 0.96 = 618.4 m$$Physics

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