Question

Which projectile spends more time in the air, the one fired from 30∘ or the one fired from 60∘?

Answer 1

time of flight is given by $T=\frac{2usin\theta}{g}$

here , u is the initial velocity , $\theta$ is the angle of projection of projectile with horizontal and g is acceleration due to gravity.

so, if u is constant then, time of flight is directly proportional to $sin\theta$

case 1 :- $\theta=30^{\circ}$
so, sin30° = 1/2 = 0.5

case 2 :- $\theta=60^{\circ}$
so, sin60° = √3/2 = 0.866

here we can see that sin60° > sin30°
so, time of flight in case of 60° projection angle is greater than 30°.

hence, projectile spends more time in the air, when angle of projection is 60°

Answer 2

V^2/gsin2Θ, well sin 60° and sin 120° are just the same ...it means that shot length is the same and but we know that D is also equal to Vcos Θ*t well:

the greater the angle, the less VcosΘ it means that the flying time is higher for an angle of 60° and it is the best formula for maximum height.