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Nikolay [14]
3 years ago
5

A ball is projected upward at time t = 0.0 s, from a point on a roof 90 m above the ground. The ball rises, then falls and strik

es the ground. The initial velocity of the ball is 36.2 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to
Physics
1 answer:
aivan3 [116]3 years ago
7 0

Answer:

The time when the ball strikes the ground is closest to  t_t  = 9.4 \  s

Explanation:

From the question we are told that

  The time of projection is t = 0.0 s

   The  distance of the point  from the ground  is  d =  90 \ m

    The  initial velocity of the ball is  v _i  =  36 .2 \ m/s

generally the time required to reach maximum height is  

      t_r =  \frac{g}{v}

Where is the acceleration due to gravity  with value  g = 9.8 \ m/s^2

Substituting values

        t_r =  \frac{36.2}{9.8}

        t_r =  3.69 s

when returning the time and velocity at the roof level is  t =  3.69 s and  u = 36.2 m/s this due to the fact that  air resistance is negligible

   The final velocity at which it  hit the ground is

      v_f^2  =  u^2  + 2ag

So  

    v_f  =  \sqrt{ u^2  + 2gs}

substituting values

    v_f  =  \sqrt{ 3.69^2  + 2* 9.8 * 90}

     v_f  =  55.45 \ m/s

The time taken for the ball to move from the roof level to the ground is  

     t_g  =   \frac{v-u}{a}

substituting values

    t_g  =   \frac{55.45 -36.2}{9.8}

     t_g  =  1.96 \ s

The total time for this travel is  

    t_t  = t_g + 2 t_r

     t_t  = 1.96 + 2(3.69)

      t_t  = 9.4 \  s

 

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