Acrow is flying horizontally with a constant speed of 2.70 m/s when it releases a clam from its beak. the clam lands on the rocky beach 2.10 s later. just before the clam lands, what is (a) its horizontal component of velocity, and (b) its vertical component of velocity? (c) how would your answers to parts (a) and (b) change if the speed of the crow were increased? explain.

a) 2.70 m/sec

b) 20.6 m/sec

c) As the speed of crow increases the horizontal component of speed of the clam will also increases but there is no change in the vertical component of speed.

Given :

Speed = 2.70 m/sec

Solution :

a) There is no gravity or no other acceleration in horizontal direction. Therefore velocity in horizontal direction is,

b) To find velocity in vertical direction we can use kinematics

---- (1)

Where,

,

t = 2.10 sec

Put the values of in equation (1),

c) As the speed of crow increases the horizontal component of speed of the clam will also increases but there is no change in the vertical component of speed.

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Part a)

in horizontal direction there is no gravity or no other acceleration

so in horizontal direction the speed of clam will remain same

Part b)

In vertical direction we can use kinematics

part c)

if the speed of crow will be increased then the horizontal speed of the clam will also increase but there is no change in the vertical speed

40 J

Explanation:

Work = change in energy

W = 1/2 k (Δx)²

First we need to find the value of k.

F = kΔx

10 N = k (0.04 m)

k = 250 N/m

Solving for work

W = (250 N/m) (0.40 m)²

W = 40 J