Physics question: A bird is flying with a speed of 1.88 m/s. There is a reservoir that is 3.89m below.?

Question

The bird is also flying at an angle of 15.32o below the horizontal. She has a fish in her claws and drops it. What will be the fish's VECTOR velocity or final velocity when it splashes?

Answer 1

The vertical component of the speed is 1.88(sin15.32). That is the same vertical component of the speed of the fish at the point where the bird had dropped it. So now we have its initial vertical component of its velocity , i.e. 1.88sin(15.32). We also have the height of the vertical drop which is 3.89meters.

The appropriate formula to use is: v^2-u^2=2as, where v is the unknown final vertical component of its velocity, u the initial vertical component of its velocity of
1.88sin15.32, a is the acceleration of gravity of
9.8m/s/s, and s the given vertical drop of 3.89meters.

Substitute the values in the formula v^2-u^2=2as:

v^2-1.88sin15.32=2*9.8*3.89
v^2=2*9.8*3.89+1.88sin15.32
=76.2+1.88sin15.32
v=(76.2+1.88sin15.32)^1/2 (You know how to compute
this, right?)

But that's the vertical component of its velocity; its vector velocity will be v/sin15.32 which is also 15.32 degrees from the horizontal.

Answer 2

Need to know altitude of lake above sea level, barometric pressure that instant, temperature of the air, and, by the way, the vector of the 1.88 m/s. Air speed? or speed over ground?

Answer 3

Just determine the additional kinetic energy accumulated from the drop and add it to the initial kinetic energy and solve for the velocity