A seagull diving towards a stone beach at an angle of 60 degrees to the vertical releases a clam from a height of 100 meters. The clam hits the beach seconds later. To the nearest tenth of a 2.32 m/s what was its speed when it was released?

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Answer 1 · 2016-07-01T14:55:57

88.71 m/s

Use

square of final speed = square of initial speed + 2 g (distance),

in the vertical direction..

If v is the speed of the bird, flying at #60^o# to the vertical,

the magnitude of the component velocity, in the vertical direction, is

#v cos 60^o=v/2#.

I assume that the clam hits the ground at 2.32 m/s, in the vertical

direction, nearly. Then,

#(v/2)^2=2.32^2+2(9.81)(100) #. Solving,

#v = 88.71 #m/s