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A 0.52 kg projectile is fired into the air from the top of a 6.31 m cliff above a valley. Its initial velocity is 9.81 m/s at 39◦ above the horizontal. Cliff x 39 9 ◦ .81 m/s 6.31 m How long is the projectile in the air? The acceleration due to gravity is 9.8 m/s 2 . Answer in units of s.

Respuesta :

Answer:

The projectile flies 1.93 seconds in the air.

Explanation:

Consider the vertical motion of projectile,

      We have equation of motion s = ut + 0.5 at²

              Initial velocity, u = 9.81 x sin 39 = 6.17 m/s

              Acceleration, a = -9.81 m/s²

              Displacement is zero when the shell reaches back to ground, s = -6.31 m

              Time, t =?

      Substituting,

              s = ut + 0.5 at²

              -6.31 = 6.17  x t + 0.5 x (-9.81) x t²

             4.905 t² - 6.17 t - 6.31 = 0

             [tex]t=\frac{-(-6.17)\pm \sqrt{(-6.17)^2-4\times 4.905\times (-6.31)}}{2\times 4.905}\\\\t=\frac{6.17\pm \sqrt{161.87}}{9.81}\\\\t=\frac{6.17\pm 12.72}{9.81}\\\\t=1.93s\texttt{ or }t=-0.67s[/tex]

Negative time is not possible,

The projectile flies 1.93 seconds in the air.