Question

Body is projected with velocity you search that its horizontal range and maximum vertical height at the same the maximum height is

Answer 1

Ans -: One possible solution to this problem seems to be manipulation of trigonometric ratios. We know that―

In a projectile motion;

Range(R)=u2sin(2a)gRange(R)=u2sin(2a)g and

Height(H)=u2sin2(a)2gHeight(H)=u2sin2(a)2g

where, uu is the initial velocity or velocity of projection, aa is the angle of projection and gg is the acceleration due to gravity.

Using the identify, sin(2a)=2sin(a)cos(a)sin(2a)=2sin(a)cos(a)in the range formula and dividing by the height formula, we get the relation― tan(a)=4HRtan(a)=4HR …(1)

We also know an identify that― sin(2a)=2tan(a)1+tan2(a)sin(2a)=2tan(a)1+tan2(a)

Substitution for tan(a)tan(a) from (1), willl give,

sin(2a)=8RHR2+16H2sin(2a)=8RHR2+16H2

Putting this result back in the formula for range will give―

R=(u2)∗8RH(R2+16H2)∗gR=(u2)∗8RH(R2+16H2)∗g

=>u=(R2+16H2)∗g8H−−−−−−−−−√=>u=(R2+16H2)∗g8H (Answer)

Don't forget to try yourself. You might get some other simple relation.

Hope this helps.

THANKS FOR READING. :-)