Question

If the archerfish spits its water 30 degrees from the horizontal aiming at an insect 1.2 m above the surface of the water, how fast must the fish spit the water to hit its target? The insect is at the highest point of the trajectory of the spit water. Use g = 10 m/s2.

Answer 1

Answer:

The speed is  $v = 9.8 \ m/s$

Explanation:

From the question w are told that

    The angle  made is $\theta = 30^o$

     The distance  above the surface of the water is  $h_{max} = 1.2 \ m$

     The  value of  $g = 10 \ m/s^2$

   

The maximum height attained by the fish is mathematically evaluate as

       $h_{max} = \frac{v^2 sin ^2 \theta }{2g }$

Making v which is the speed of the fish the subject of the formula

      $v = \sqrt{ \frac{2gh_{max}}{ sin^2 \theta } }$

  substituting values

     $v = \sqrt{ \frac{2*10 *1.2 }{ [sin (30)]^2 } }$

     $v = 9.8 \ m/s$