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An airplane takes off a runway at a constant speed of 49m/s at constant angle 30 to the horizontal

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Complete Question

An airplane takes off a runway at a constant speed of 49 m/s at constant angle 30 to the horizontal.How high (in meters )  is the airplane above the ground 13 seconds after takeoff?

Answer:

The height  is [tex]H = 318.5 \ m[/tex]

Explanation:

From the question we are told that

   The speed at which the plane takes off is  [tex]u = 49 \ m/s[/tex]

      The angle at which it takes off is  [tex]\theta = 30 ^o[/tex]

        The time taken is [tex]t = 13 s[/tex]

The vertical distance traveled is  mathematically represented as

          [tex]H = u sin \theta t[/tex]

Substituting values  

         [tex]H = (49) * sin (30) *13[/tex]

        [tex]H = 318.5 \ m[/tex]