Question

An object is thrown straight up into the air with an initial speed of 8 m/s, and reaches a greatest height of 15 m before it falls back to the ground. a. What height is the object thrown from? b. How long does it take for the object to hit the ground?

Answer 1

Answer:

The value is $T_t = 2.5659 \ s$    

Explanation:

From the we are told that  

         The initial speed of the object is  $u = 8 \ m/s$

         The greatest height it reached is  $h = 15 \ m$

Generally from kinematic equation we have that

      $v^2 = u^2 + 2gH$

At maximum height v  =  0 m/s

So

      $0^2 = 8^2 + 2 * - 9.8 * H$

=>    $H = 3.27 \ m$

Here H is the height from the initial height to the maximum height

So the initial height is mathematically represented as  

      $s = h - H$

=>    $s = 15 - 3.27$

=>    $s = 11.73 \ m$

Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

            $v = u + (-g) t$

At maximum height v  =  0 m/s

           $0 = 8 - 9.8t$

=>         $t = 0.8163 \ s$

Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

       $h = ut_1 + \frac{1}{2} gt_1^2$

Here the initial velocity is  0 m/s given that its the velocity at maximum height

Also  g is positive because we are moving in the direction of gravity  

So

       $15 = 0* t + 4.9 t^2$

=>      $t_1 = 1.7496$

Generally the total time taken is mathematically represented as

          $T_t = t + t_1$

=>        $T_t = 0.8163 + 1.7496$

=>        $T_t = 2.5659 \ s$