Projectile Motion rock throw Help

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To determine the velocity of a rock thrown horizontally from a cliff at 25 m/s after 3 seconds, the correct approach involves treating the horizontal and vertical velocities as vector components rather than averaging them. The vertical velocity after 3 seconds, due to gravity, is 29.4 m/s. The total velocity can be calculated using the Pythagorean theorem, resulting in a magnitude of 38.6 m/s and a direction of 49.6 degrees downward. Understanding vector addition and basic trigonometry is essential for solving this type of projectile motion problem. Proper application of these principles will yield the correct results.
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If a rock is thrown horizontally from a cliff at 25 m/s, what will its velocity be (magnitude and direction) after 3.0 seconds. The answer is 38.6 m/s, and 49.6 degrees down, but I cannot figure out how to get to it.



Not sure which equations to use, I just started and I am learning on my own.



I assumed that to find the velocity, you should average the horizontal and vertical velocities together. So since it has been 3 seconds, the vertical would be 29.4 m/s, because the gravitational constant is 9.8 m/s/s, and the horizontal would remain 25 m/s. Averaged, it would be 27.2 m/s. Obviously that was incorrect. As far as direction goes, I have no idea.
 
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The horizontal and vertical velocities are the components of the total velocity. You must add them like vectors, not average them. A bit of trig will allow you to find to the magnitude and direction of that vector.
 
Oh okay, I'll look that up and see if I can get it. Thanks
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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