If R and H are the horizontal range and maximum height attained by a projectile, than its speed of projection is
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Answered by
197
Range(R)=u2sin(2a)/g
and
Height(H)=u2sin2(a)/2g
where, u is the initial velocity or velocity of projection, a is the angle of projection and g is the acceleration due to gravity.
Using the identify, sin(2a)=2sin(a)cos(a) in the range formula and dividing by the height formula, we get the relation― tan(a)=4H/R …(1)
We also know an identify that― sin(2a)=2tan(a)/1+tan2(a)
Substitution for tan(a) from (1), willl give,
sin(2a)=8RH/R2+16H2
Putting this result back in the formula for range will give―
R=(u2)∗8RH/(R2+16H2)∗g
Answer = u = √(R2+16H2)∗g/8H
and
Height(H)=u2sin2(a)/2g
where, u is the initial velocity or velocity of projection, a is the angle of projection and g is the acceleration due to gravity.
Using the identify, sin(2a)=2sin(a)cos(a) in the range formula and dividing by the height formula, we get the relation― tan(a)=4H/R …(1)
We also know an identify that― sin(2a)=2tan(a)/1+tan2(a)
Substitution for tan(a) from (1), willl give,
sin(2a)=8RH/R2+16H2
Putting this result back in the formula for range will give―
R=(u2)∗8RH/(R2+16H2)∗g
Answer = u = √(R2+16H2)∗g/8H
Answered by
108
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