Question

a projectile is launched from a cannon with an initial vertical speed of 61m/s upward and a horizontal speed of also 61 m/s. the projectile travels a horizontal distance of 759.3 m before returning back to the initial vertical position. what is the time of flight of the projectile?

Answer 1

12.45 s

Step-by-step explanation:

Given in the x-direction:

s = 759.3 m

u = 61 m/s

a = 0 m/s²

Find: t

s = ut + ½ at²

759.3 = (61) t + ½ (0) t²

t = 12.45 s