What is the Required Speed for a Successful Long Jump Shot?

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To successfully make a long jump shot, the ball must be released 1.7 meters above the ground and travel a horizontal distance of 6.03 meters to reach a basket height of 3.05 meters. Calculating the required speed involves using projectile motion equations, particularly considering launch angles of 30° and 60°. The initial attempts to find speed using basic formulas were unsuccessful, prompting a recommendation to refer to textbooks or online resources for more detailed projectile motion information. Understanding the relationship between launch angle, height, and distance is crucial for determining the necessary speed. Accurate calculations are essential for achieving a successful long jump shot.
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long jump shot is released 1.7 m above the ground, 6.03 m from the base of the basket, which is 3.05 m high. For launch angles of 30° and 60°, find the speed needed to make the basket.

_______m/s (30° launch angle)
_______m/s (60° launch angle)
 
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Show relevant equations and your attempts.
 
okay i tried using speed=meters/seconds
but it didn't work.

so i plugged in into an equation
like v=1/2gt^2
 
Go through your textbook or any site and find the information regarding the projectile motion.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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