If an arrow is fired from a bow with a perfectly horizontal velocity of 60.0 m/s and the arrow had a vertical velocity of 6m/s just before it hit the ground, how long will it take to strike the ground ?

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

the answer is kevlar

a skateboard rolling across the street - apex