What's the Correct Formula for Finding the Speed of a Thrown Ball?

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SUMMARY

The correct speed for a ball thrown horizontally from a height of 20 meters, striking the ground at a 45° angle, is 20 m/s. The formula h = (Vi^2 sin^2 alpha) / (2g) is not applicable in this scenario. Instead, the appropriate equations of motion, specifically s = s0 + ut - 0.5gt², should be utilized to determine the time of flight and initial velocity. The initial vertical velocity is zero since the ball is thrown horizontally.

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A ball is thrown horizontally from the top of a 20-m high hill. It strikes the ground at an angle of 45°. With what speed was it thrown? I thought that we could use h=(Vi^2sin^2alpha)/2g. However, using that formula I am not able to get to the correct answer which is 20m/s. What am I doing wrong?

Thanks!
 
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You know the formula s=s0+ut-0.5gt2 where s = displacement and u= initial velocity.

Using this formula and considering vertical motion, the ball hits the ground at s=0. So what is the total time of flight? (If it was thrown horizontally, its initial vertical velocity is?)
 

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