Answer:
t = 1,28 s
Explanation: This problem is a projectile motion problem.
V₀ₓ = V₀ * cosθ
Vₓ = constant all the way
Vₓ = V₀ₓ
tanθ = Vyi/Vxi that means in any point of the trajectory
tanθ = 6 / 60 ( just before touching the ground )
tanθ = 0,1 then arctan 0,1 ≈ 6⁰
sin 6⁰ = 0,1045
cos 6⁰ = 0,9945
V₀ₓ = V₀ * cosθ ⇒ V₀ = V₀ₓ / cosθ ⇒ V₀ = 60 / 0,9945
V₀ = 60,33 m/s
V₀y = V₀ * sin θ ⇒ V₀y = 60,33 * 0,1045 ⇒ V₀y = 6,30 m/s
Vy = V₀y - g * t
At maximum y Vy = 0 ( the middle of the trajectory)
g*tm = V₀y ⇒ tm= V₀y / g
tm = 6,30 / 9,8
tm = 0,64 s ( time to reach maximum y )
Then the time of fligh is twice 0,64 s
t = 0,64 * 2
t = 1,28 s
