# 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?

Jul 1, 2016

88.71 m/s

#### Explanation:

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

$v \cos {60}^{o} = \frac{v}{2}$.

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

direction, nearly. Then,

${\left(\frac{v}{2}\right)}^{2} = {2.32}^{2} + 2 \left(9.81\right) \left(100\right)$. Solving,

$v = 88.71$m/s