Question

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 strikes 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

Answer 1

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  

    $t_t = t_g + 2 t_r$

     $t_t = 1.96 + 2(3.69)$

      $t_t = 9.4 \ s$