Question

i want its complete solution

Answer 1

Hey. Hope this helps! It is given maximum horizontal distance which means theta is 45 degrees. Therefore range=v^2sin2theta/g.putting values of theta, range and g. We get v =root 3...
Now component along x axis =vcos theta=root 3*cos 45=root3/root2......
B option is correct

Answer 2

we know that....
range is maximum only when angle of projection is 45°
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also we know that...
Range = R= (u^2 sin2θ)/g

=> R = u^2(sin2(45°))/g

=> R= 0.3= u^2(sin90°)/g

=> R= 0.3×g= u^2

=> u^2= 0.3×10 = 3

=>u=√3m/s
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we know that....
horizontal component of velocity is .
ux= ucosθ
{where ux= velocity along x-axis}

=> ux= √3 cos 45°

=> ux= √3×1/√2

=> ux= √(3/2) m/s
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so the answer will be 2 nd option...
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hope it helps you...
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