Question

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 ?

Answer 1

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