Determining Speed given Horizontal and Vertical Distances

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SUMMARY

The discussion focuses on calculating the initial speed at which a girl threw a ball that landed 21.0 meters high and 20.0 meters horizontally from the launch point. The relevant kinematic equations include vfy = viy + ay(delta t), vfy^2 = viy^2 + 2ay(delta y), and v = sqrt(vx^2 + vy^2). The calculated initial speed is 22.5 m/s, derived from a straightforward application of these equations. Participants emphasize the importance of showing work for better guidance in solving the problem.

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Homework Statement


A girl throws a ball high into the air, and at its highest point it lands on a roof, where it gets stuck. You estimate the ball landed at a height of 21.0m from the ground and a horizontal distance of about 20.0m from where it was thrown. What is the speed that the girl threw the ball at?

Homework Equations


vfy = viy + ay(delta t)
vfy^2 = viy^2 + 2ay(delta y)
vfx = vix + ax(delta t)
x = vix (delta t) + 1/2(ax)(delta t)^2
y = viy (delta t) + 1/2 (ay) (delta t)^2
v = sqrt (vx^2) + (vy^2)

The Attempt at a Solution


It would take an enormous amount of space to replicate my work (and this would honestly be no better than spam), but the answer I found was 22.5 m/s. Please help me.
 
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It wouldn’t help your understanding much for us to just tell you how to do it, so I’m afraid you will have to show us your attempts so we can subtly suggest and guide.

However, I can say it shouldn’t take an enormous amount of work. There is a straight forward approach with a couple of those equations that takes only a little work.
 

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