Question

A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 30 m/s. if the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10 m/s, how fast would the stone be traveling just before it hits the ground

Answer 1

Answer:

The velocity just before hitting the ground is $v_f = 30 m/s$

Explanation:

From the question we are told that

    The initial speed is  $u = 10 m/s$

    The final speed is  $v = 30 \ m/s$

From the equations of motion we have that

      $v^2 =u^2 + 2as$

Where s is the distance travelled which is the height of the cliff

  So making it the subject of the the formula  we have that

        $s = \frac{v^2 - u^2 }{2a}$

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

       So

                  $s = \frac{30^2 - 10^2 }{2 * 9.8 }$

                  $s = 40.8 \ m$

Now we are told that was through horizontally with a speed of

      $v_x =10 m/s$

Which implies that this would be its velocity horizontally through out the motion

    Now it final  velocity vertically can be mathematically evaluated as

            $v_y = \sqrt{2as}$

Substituting values

             $v_y = \sqrt{(2 * 9.8 * 40.8)}$

             $v_y = 28.3 \ m/s$

The resultant final velocity is mathematically evaluated as

       $v_f = \sqrt{v_x^2 + v_y^2}$

Substituting values

       $v_f = \sqrt{10^2 + 28.3^2}$

       $v_f = 30 m/s$