Question

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

Answer 1

Answer: 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.

Answer 2

Answer:

∴ the angle for greatest distance is 90°, where Vy = 0

Explanation:

maximum height of a projected object is highest vertical position of the trajectory object which depends on the initial velocity

formula for maximum height

H= u² sin²θ/2g

if θ = 90°, then sin²90° = 1

∴ the angle for greatest distance is 90°, where Vy = 0

for horizontal distance,

the range of a projectile is the horizontal distance.

R = u² sin2θ/g

if θ = 45°, then 2θ  = 90°

∴sin 90°= 1