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

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To find the speed of a ball thrown horizontally from a 20-meter-high hill that strikes the ground at a 45° angle, the correct approach involves analyzing both horizontal and vertical motions. The formula h = (Vi^2 sin^2 alpha) / (2g) is not suitable for this scenario since the initial vertical velocity is zero. Instead, using the equation s = s0 + ut - 0.5gt^2 helps determine the time of flight based on vertical displacement. The total time of flight can be calculated from the height and gravitational acceleration, leading to the conclusion that the initial speed required for the ball to hit the ground at the specified angle is indeed 20 m/s. Understanding the relationship between vertical and horizontal components is crucial for solving this problem accurately.
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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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