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A projectile is shot in the air from ground level with an initial velocity of 560 m/sec at an angle of 30° with the horizontal. (Round your answers to two decimal places.) (a) At what time (in seconds) is the maximum range of the projectile attained? s (b) What is the maximum range (in meters)?

Answer :

Answer:

a)   t = 57.14 s

b)    x = 27711.4 m

Explanation:

This is a missile throwing exercise

a) They ask us for the time to the maximum reach, this corresponds to when it reaches the ground y = 0, let's use

          y =  [tex]v_{oy}[/tex] t - ½ g t²

Let's use trigonometry to find the vertical initial velocity

         sin θ = v_{oy} / v₀

         v_{oy} = v₀ sin θ

           

we substitute

         y = v₀ sin θ t - ½ g t²

since the height is zero

          0 = t (v₀ sin θ - ½ g t)

This equation has two solutions

*  t = 0 which corresponds to the moment of launch

*

         v₀ sin 30 - ½ g t = 0

        t = v₀ sin 30    2/g

let's calculate

         t = 560 sin 30  2 / 9.8

         t = 57.14 s

b) with this time we can calculate the distance traveled

          x = v₀ₓ t

     

let's use trigonometry for velocity

          cos θ = v₀ₓ / v₀

          v₀ₓ = v₀ cos 30

we substitute

         x = v₀ cos 30     t

let's calculate

         x = 560 cos 30 57.14

         x = 27711.4 m

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