What is the speed of the ball just before it strikes the ground?

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SUMMARY

The problem involves a ball thrown from a height of 46 meters at an angle of 47 degrees with an initial speed of 23 m/s. The ball travels a horizontal distance of 82.0127 meters before hitting the ground, with gravity acting at 9.8 m/s². The correct speed of the ball just before impact is calculated to be approximately 56.16 m/s, although there was confusion regarding the maximum height and time calculations. Accurate diagramming and understanding of projectile motion are essential for solving this problem correctly.

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A ball is thrown from the top of a building
upward at an angle of 47 ◦ to the horizontal
and with an initial speed of 23 m/s, as in
the figure. The ball is thrown at a height of
46 m above the ground and hits the ground
82.0127 m from the base of the building.
The acceleration of gravity is 9.8 m/s2 .

What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s.


I have tried calculating the max height (14.4363m) and the time @ max height (1.71644s) then doing vf-=vi-gt => vf=-9.8(1.71644)=-16.8211 m/s (wrong)

not sure what else to try...
 
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I'm not sure how you managed to get max height at 14.4363m when the ball is ''thrown at a height of 46 m above the ground.'' Try redrawing your diagram, I can show you how to get answer, just tell me if the answer is v=56.16m/s otherwise I will have to recheck my drawing.

Best of luck
 
the answer is not v= 56.16 m/s

thanks though
 

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