Somewhere in the vast flat tundra of planet tehar, a projectile is launched from the ground at an angle of 60 degrees. it reaches the maximum height of 15 m. the acceleration due to gravity is 30 m/s2. find the horizontal range of the projectile in meters.

R = 0.0503 m

Explanation:

This is a projectile launching exercise, to find the range we can use the equation

       R = v₀² sin 2θ / g

How we know the maximum height

      ² =² - 2 g y

      = 0

      = √ 2 g y

      = √ 2 9.8 / 15

      = 1.14 m / s

Let's use trigonometry to find the speed

    sin θ = / vo

    vo = / sin θ

    vo = 1.14 / sin 60

    vo = 1.32 m / s

We calculate the range with the first equation

     R = 1.32² sin(2 60) / 30

    R = 0.0503 m

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