Two projectiles are thrown with the same initial speed, one at an angle θ with respect to the level ground and the other at angle 90° − θ. Both projectiles strike the ground at the same distance from the projection point. Are both projectiles in the air for the same length of time?

Answer: Height is different for both .The time in air are different for both.Step-by-step explanation:Given that initial velocity is sameLets take initial velocity = u One at  an angle θ.other at angle 90° − θ.Also given that their range are same[tex]R_1=R_2[/tex][tex]R_1=\dfrac{u^2sin2\theta }{g}[/tex][tex]R_2=\dfrac{u^2sin2(90-\theta)}{g}[/tex][tex]R_2=\dfrac{u^2sin(180-2\theta)}{g}[/tex]We know thatsin(180° − θ)=sin θSo[tex]R_2=\dfrac{u^2sin2\theta }{g}[/tex]Height in the air [tex]h_1=\dfrac{u^2sin^2\theta}{2g}[/tex][tex]h_2=\dfrac{u^2sin^2(90-\theta)}{2g}[/tex]We know thatsin(90° − θ)=cos θ[tex]h_2=\dfrac{u^2cos^2\theta}{2g}[/tex]          From above we can say that height is different for both .Time:[tex]T_1=\dfrac{2usin\theta}{g}[/tex][tex]T_2=\dfrac{2usin(90-\theta)}{g}[/tex]sin(90° − θ)=cos θ[tex]T_2=\dfrac{2ucos\theta}{g}[/tex]The time in air are different for both.