Question

The range of a projectile, launched with a speed u at an angle theta with the horizontal is given by R=2u^2 sin theta cos theta/g = u^2 sin 2theta /g. Find the angle of projection for maximum range R for given u.​

Answer 1

Answer:

45° hope it helps

Explanation:

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Answer 2

Answer:

Hey there!

▶ Ans is angle,tita= 45°

Explanation:

For angle 45°

R = u^2 sin2tita/g

for tita=45°

R= u^2 sin2(45°)/g= u^2 sin90/g=u^2/g

thus maximum range for tita =45° is R= u^2/g