Question

A person throws a pumpkin at a horizontal speed of 4.0 — off a cliff. The pumpkin travels 9.5 m horizontally

Answer 1

Complete Question

A person throws a pumpkin at a horizontal speed of   4.0 m/s off a cliff. The pumpkin travels 9.5m horizontally before it hits the ground. We can ignore air resistance.What is the pumpkin's vertical displacement during the throw? What is the pumpkin's vertical velocity when it hits the ground?

Answer:

The  pumpkin's vertical displacement  is  $H = 27 .67 \ m$

The  pumpkin's vertical velocity when it hits the ground is  $v_v__{f}} = 23.298 \ m/s$

Explanation:

From the question we are told that

   The  horizontal speed is  $v_h = 4 m/s$

    The horizontal distance traveled is  $d = 9.5 \ m$

The horizontal distance traveled is mathematically represented as

           $S = v_h * t$

Where t is the time taken

substituting values

          $9.5 = 4 * t$

   =>     $t = \frac{9.5}{4}$

            $t = 2.38 \ sec$

Now the vertical displacement is mathematically represented as

        $H = v_v t + \frac{1}{2} a_v t^2$

now the vertical velocity before the throw is  zero

    So

          $H = 0 + \frac{1}{2} (9.8) * (2.375)^2$

          $H = 27 .67 \ m$

Now the final vertical velocity  is mathematically represented as

          $v_v__{f}} = v_v + at$

  substituting values

             $v_v__{f}} = 0 + (9.8)* (2.375)$

            $v_v__{f}} = 23.298 \ m/s$