Question

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?

Answer 1

Answer:Check the attached

Explanation:

Answer 2

Answer:

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

          $v_{oy}$ = 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

            $v_{y}$ = 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