Solving for Initial Velocity in Projectile Motion

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To solve for the initial velocity in projectile motion, the maximum horizontal distance of 136 m is used along with the acceleration due to gravity of 9.8 m/s². The range equation is applied to find the initial velocity, leading to the calculation of 30.7 m/s. This velocity is then used to determine how high the ball can be thrown vertically, assuming the same speed is maintained. The angle for maximum range is confirmed to be 45 degrees, which simplifies the calculations. The discussion concludes with a successful determination of the initial velocity needed for vertical throw calculations.
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A man can throw a ball a maximum horizontal distance of 136 m on a level field. The acceleration of gravity is 9.8 m/s^2. How far can he throw the same ball vertically upward from the ground? (Assume that the ball is thrown from the ground and that his muscles give the ball the same speed in each case.) Answer in units of m.

My work: I was trying to use the range equation to solve for vi so 136=v^2sin(2theta)/g ok so 136x9.8= 1332.8=v^2sin(2theta) , I am obviously missing something what is it? thanks
 
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What is the angle \theta for maximum range? Assume no air resistance.

When you obtain the velocity, then apply it to the vertical case.
 
45 degrees and sin(45) is sqrt(2)/2 so really its 1332.8=v^2(sqrt(2)) , which for v i got 30.69905402 thank you, i didnt think of that
 
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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