R=Vi2(sin 2θ)/g
At given condition:
Maximum range= Rmax= vi2θ/g.
For the maximum range, sin2θ=1 which is possible if we put the angle as 45 degrees but to get half of the maximum range we should take θ such that sin2θ is equal to 0.5. It is known that sin 30=0.5 so if we put θ as 15 we will get sin2(15)=sin 30=0.5.
As per the explanation, this is incorrect.
R=Vi2(sin 2θ)/g
At given condition:
Maximum range= Rmax= vi2θ/g.
For the maximum range, sin2θ=1 which is possible if we put the angle as 45 degrees but to get half of the maximum range we should take θ such that sin2θ is equal to 0.5. It is known that sin 30=0.5 so if we put θ as 15 we will get sin2(15)=sin 30=0.5.
As per the explanation, this is incorrect.
R=Vi2(sin 2θ)/g
At given condition:
Maximum range= Rmax= vi2θ/g.
For the maximum range, sin2θ=1 which is possible if we put the angle as 45 degrees but to get half of the maximum range we should take θ such that sin2θ is equal to 0.5. It is known that sin 30=0.5 so if we put θ as 15 we will get sin2(15)=sin 30=0.5.
As per the explanation, this is incorrect.
R=Vi2(sin 2θ)/g
At given condition:
Maximum range= Rmax= vi2θ/g.
For the maximum range, sin2θ=1 which is possible if we put the angle as 45 degrees but to get half of the maximum range we should take θ such that sin2θ is equal to 0.5. It is known that sin 30=0.5 so if we put θ as 15 we will get sin2(15)=sin 30=0.5.
R=Vi2(sin 2θ)/g
At given condition:
Maximum range= Rmax= vi2θ/g.
For the maximum range, sin2θ=1 which is possible if we put the angle as 45 degrees but to get half of the maximum range we should take θ such that sin2θ is equal to 0.5. It is known that sin 30=0.5 so if we put θ as 15 we will get sin2(15)=sin 30=0.5.