Question

Pad? 11:12 PM * 74%. a flipitphysics.com Unit 3: Checkpoint Homework Problems Homework: Iwo Dimensional DynamiCS Print Assignment View Deadline: 100% until Friday, June 15 at 11:59 PM Interactive Example Arrow What A Shot! Standard Exercise Ball Throw #1 Interactive Example Balloon Standard Exercise What A Shot! Standard Exercise Projectile Standard Exercise Ball Throw #2 Standard Exercise Airplane Standard Exercise Roof Toss Standard Exercise Cliff Toss The archerfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fish's mouth. Suppose the archerfish squirts water with an initial speed of 2.35 m/sec at an angle of 24.3° above the horizontal. When the stream reachs the beetle a height h avoe the water's surface it is moving horizontal Standard Exercise Bouncing Ball

1) what is the vertical component of the waters initial velocity?

2) how long does the beetle have to react?

3) how high is the beetle above the fish?

What is the horizontal distance between the fish and the beetle?

Answer 1

Solution:

Initial velocity =v = 2.35 m/s

Direction = $\theta$ =24.3 degrees

1) Initial vertical velocity = v_y = v sin $\theta$ = 2.35 sin24.3 = 0.97 m/s

2) Beetle's time to react = time taken by water to reach the insect

Since only horizontal component reaches it (given) , the vertical component of velocity is 0 there.

vfy = viy - gt

=> 0 - v sin $\theta$ -gt

=> t = v sin $\theta$ /g = 2.35 sin24.3 /9.81 = 0.9671 /9.81 = 0.09858 s

= 0.1 s

3) height above the fish = h = ( v sin $\theta$ )² /2g = (0.9671)² / 19.6 = 0.048 m

= 0.05 m

4) Horizontal distance = d = v cos $\theta$ * t = (2.35 cos 24.3) * ( 0.09858 ) = 0.21 m