What angle would you use for the greatest distance when the projectile leaves at zero height above ground g

45 degree

Explanation:

Since Range = Ucosø × T

Where T = total time.

Range = distance covered

T = 2usinø/ 2g

If you substitute T into the range formula, you will get

R = (Ucosø×2Usinø) / 2g

But in trigonometry, 2sinøcosø = sin2ø

Substitute it into the formula

R = Usin2ø/ 2g

If ø = 45 degree

R = Usin(2 × 45)/2g

R = Usin90 / 2g

But sin 90 = 1

Therefore,

Range R = U/2g

Therefore, 45 degree angle would be use for the greatest distance when the projectile leaves at zero height above ground.

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