Question

A crow is flying horizontally with a constant speed of 3.00 m/s when it releases a clam from its beak. The clam lands on the rocky beach 2.40s later. Consider the moment just before the clam lands. (Neglect air
resistance.)

Answer 1

(Note: you did not complete the question and posted. I am still attempting because I have solved the similar kind of question earlier as well.)

Solution: Given that,

speed of the crow = v(i) = 3 m/s

time taken by the clam = t = 2.4 s

(a)We need to find the horizontal component of velocity. Let it be v(ix).

As the crow is flying horizontally, so angle = = 0

We have, Intial velocity of the crow = vi = 3 m/s

So the horizontal compotent of its velocity will be:

v(ix) = vi cos = 3 x cos(0) = 3 x 1 = 3 m/s

Hence, v(ix) = 3 m/s

(b)We need to find the vertical component of velocity.Let it be v(y).

v(iy) will be, v(iy) = vi x sin = 3 x sin0 = 0

We know from the equation of motion that,

v = u + at

In our case, v = v(y) , v = v(iy) , a = -g = -9.8 and t = 2.4 s

v(y) = v(iy) + (-g)t = 0 - 9.8 x 2.4 = 23.52 m/s

Hence, v(y) = 23.52 m.s

(c) We need to find from what height the clam was released. Let it be H

Again from eqn of motion we know that,

s = s0 + ut + 1/2 at²

In our case, s = 0 ; s0 = H , u = v(iy)=0 and t = 2.4s, a = -g = -9.8

0 = H + 0 + 1/2 (-g) t² => H = 0.5 x 9.8 x (2.4)² = 28.22 meters

Hence, H = 28.22 meters