CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
526.3 m
loader
C
486.6 m
loader
D
None of these
loader

Solution

The correct option is D None of these
time 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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image