Calculating the Speed of a Seagull Diving for Clams in 3.32 Seconds

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The discussion focuses on calculating the speed of a seagull diving for clams, specifically determining the velocity at which the clam was released. The seagull releases the clam from a height of 100 meters at a 60-degree angle to the vertical, and the clam impacts the beach after 3.32 seconds. Using the kinematic equation v = v_0 + a*t, participants derive the vertical and horizontal components of the velocity, ultimately leading to the calculation of the horizontal speed (vx) required for the final answer.

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1. 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 3.32 seconds later. To the nearest tenth of a m/s what was its speed when it was released?



2. v=v_0 + a*t
 
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If v is the velocity of seagull when it releases the clam, find the horizontal and vertical components of v.
Height is given, time is given and g is known.
Use the proper kinematic equation to find vy.
From that you can find vx.
required answer is vx*t.
 
rl.bhat said:
If v is the velocity of seagull when it releases the clam, find the horizontal and vertical components of v.
Height is given, time is given and g is known.
Use the proper kinematic equation to find vy.
From that you can find vx.
required answer is vx*t.

Thanks, I think that I got it now
 

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