A bullet is shot from shoulder height (1.50m)at an angle 60.0°above the horizontalwith an initial speed of 400m/s. Neglect air resistance. a. How much time elapses before the bullet hits the ground?b. How far does the bullet travel horizontally?c. What is the bullet’simpact anglewhen landing on the ground?

a)   t₁ = 70,717 s , b)  x = 14143.4 m , c) θ = -60.02º

Explanation:

We will work on this problem using kinematics,

a) time to reach the ground, let's look for the initial speed

          = v₀ sin.60

          v₀ₓ = v₀ cos 60

          v_{oy} = 400 sin. 60 = 346.41 m / s

          v₀ₓ = 400 cos 60 = 200 m / s

Let's use the projectile launch equation

           y = y₀ + v_{oy} t - ½ g t²

Upon reaching the ground y = 0

            0 = 1.5+ 346.41 t - ½ 9.8 t²

             4.9 t² - 346.41 t - 1.5 = 0

             t² - 70.696 t - 0.306 = 0

We solve the second degree equation

            t = [70.696 ±√ (70.696² + 4 0.306)] / 2

            t₁ = 70,717 s

            t₂ = -0.0085 s

We take the positive time

b) let's use the equation

              x = v₀ₓ t

              x = 200 70,717

              x = 14143.4 m

c) to find the angle let's look for the velocity at the point of impact, the velocity on the x-axis is constant

             vₓ = v₀ₓ = 200 m / s

The speed on the y axis

            = v_{oy}  - g t

            v_{y} = 346.41 - 9.8 70.717

            v_{y} = -346.62 m / s

Let's use trigonometry

           tan θ = v_{y} / vₓ

           θ = tan⁻¹ v_{y} / vₓ

           θ = tan⁻¹ (-346.62 / 200)

           θ = -60.02º