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A startled armadillo leaps upward, rising 0.587 m in the first 0.193 s. (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of 0.587 m? (c) How much higher does it go? Use g=9.81 m/s2.

Respuesta :

Answer:

a) Initial speed as it leaves the ground is 3.99 m/s

b) Speed at the height of 0.587 m is 2.10 m/s

c) Height reached is 0.81 m

Explanation:

a) We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = ?

        Acceleration, a = -9.81 m/s²  

        Time, t = 0.193 s

        Displacement, s = 0.587 m      

     Substituting

                      s = ut + 0.5 at²

                      0.587 = u x 0.193 + 0.5 x -9.81x 0.193²

                      u = 3.99 m/s

    Initial speed as it leaves the ground is 3.99 m/s

b) We have equation of motion v = u + at

     Initial velocity, u = 3.99 m/s

     Final velocity, v = ?

     Time, t = 0.193 s

     Acceleration, a = -9.81 m/s²

   Substituting

                      v = u + at  

                      v = 3.99 + -9.81 x 0.193

                      v = 2.10 m/s

    Speed at the height of 0.587 m is 2.10 m/s

c) We have equation of motion v² = u² + 2as

Initial velocity, u = 3.99 m/s  

Acceleration, a = -9.81 m/s²  

Final velocity, v = 0 m/s  

Substituting  

v² = u² + 2as

0² = 3.99² + 2 x -9.81 x s

s = 0.81 m

Height reached is 0.81 m

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