Question

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.

Answer 1

Answer:

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

      $v_{f}$² =$v_{oy}$² - 2 g y

      $v_{f}$= 0

      $v_{oy}$ = √ 2 g y

      $v_{oy}$ = √ 2 9.8 / 15

      $v_{oy}$ = 1.14 m / s

Let's use trigonometry to find the speed

    sin θ = $v_{oy}$ / vo

    vo = $v_{oy}$ / 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