The maximum range of projectle is R° if projected at angle of 45°. It'll be 300% of Range of projectile if projected at angle? a) 30° b) 45° c) 60° d) 76° Correct answer with good explanation

The maximum range is 300% of the range of the projectile is projected at an angle of 9.74°.

None of the options are correct.

Explanation:

Normally, ignoring air resistance, for projectile motion, the range (horizontal distance travelled) of the motion is given as

R = (u² sin 2θ)/g

where

u = initial velocity of the projectile

θ = angle above the horizontal at which the projectile was launched

g = acceleration due to gravity = 9.8 m/s²

The range is maximum when θ = 45°

R when θ = 45° is

R = (u² sin 90°)/g = (u²/g)

We are then told that this maximum range is 300% of the value obtainable for the range at a particular angle

Maximum range = 3R

(u²/g) = 3(u² sin 2θ)/g

Sin 2θ = (1/3)

2θ = sin⁻¹ (1/3) = 19.47°

θ = (19.47°/2) = 9.74°

Hope this Helps!!!