Question

You are woken up in the morning on the farm by a rooster. The rooster is located 30 meters away on top of the 5 meter barn house. If you throw your alarm clock at an angle of 53 degrees at the rooster, so that it hits the rooster on the way up, with what velocity do you need to throw your alarm clock

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity is  $v = 18.73 \ m/s$

Explanation:

From the question we are told  that

     The distance of the rooster from farm house is  $d = 30 \ m$

       The height of the rooster from the ground is  $h = 5 \ m$

       The angle at which the alarm clock is thrown is $\theta = 53 ^o$

       

Let the velocity at which the alarm clock is thrown be  v

  So the horizontal component  of v is mathematically represented  as  

    $v_h = v cos \theta$

The distance covered by the alarm clock toward the horizontal direction at velocity v is

         $d = v_h$

=>     $d = v_h = vcos \theta * t$

=>       $vcos \theta * t = 30$

substituting for  $\theta$

           $v (cos (53)) * t = 30$

=>       $v * t = 50$

Considering the motion of the alarm clock in the vertical direction

       So the vertical component  of v is mathematically represented  as

    $v_v = vsin \theta$

The distance covered in the vertical direction is mathematically evaluated  as follows

From the equation of motion we have

      $h = vsin\theta * t - \frac{1}{2} g t^2$

      $v* t sin\theta - 4.9t^2 = 5$

Recall  $v * t = 50$

So  

      $50 \ sin \theta -4.9t^2 = 5$

 substituting for  $\theta$

     $50 \ sin (53) -4.9t^2 = 5$

=>  $t =\sqrt{7.1289}$

      $t =2.673 \ s$

Now  

     $v * t = 50$

So

   $v * 2.673 = 50$

=>   $v = 18.73 \ m/s$