Two projectiles A and B are fired simultaneously from a level, horizontal surface. The projectiles are initially 62.2 m apart. Projectile A is fired with a speed of 19.5 m/s at a launch angle 30° of while projectile B is fired with a speed of 19.5 m/s at a launch angle of 60°. How long
it takes one projectile to be directly above the other?​

Let the point where A is launched act as the origin, so that the horizontal positions at time t of the respective projectiles are

• A : x = (19.5 m/s) cos(30°) t

• B : x = 62.2 m + (19.5 m/s) cos(60°) t

These positions are the same at the moment one projectile is directly above the other, which happens for time t such that

(19.5 m/s) cos(30°) t = 62.2 m + (19.5 m/s) cos(60°) t

Solve for t :

(19.5 m/s) (cos(30°) - cos(60°)) t = 62.2 m

t = (62.2 m) / ((19.5 m/s) (cos(30°) - cos(60°))

t ≈ 8.71 s

if the thermos has a vacuum innner, then heat through conv currents and conduction are reduced = vacuum flask

answer: strength i guess

explanation: