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:
∴ 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
