Projectile Motion Question #2 help?

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SUMMARY

The discussion focuses on solving a projectile motion problem involving a quarter tossed at a velocity of 6.4 m/s at a 60-degree angle, landing in a dish 2.1 m away horizontally. The calculated height of the shelf is confirmed to be 1.5 m. For the vertical component of the velocity just before landing, the participant initially calculated -0.93 m/s but noted the correct answer is -0.89 m/s, emphasizing the importance of precision in time calculations, which should be at least three significant figures for accuracy.

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Homework Statement


In a carnival booth, you win a stuffed giraffe if you toss a quarter over a small dish. The dish is on a shelf above the point where the quarter leaves your hand and is a horizontal distance of 2.1 m from this point. If you toss the coin with a velocity of 6.4 m/s at an angle of 60 degrees above the horizontal, the coin lands in the dish. You can ignore air resistance.

a) What is the height of the shelf above the point where the quarter leaves your hand?
b) What is the vertical component of the velocity of the quarter just before it lands the dish?




Homework Equations


x=(V0cosa0)t
y=(V0sina0)t - 1/2gt^2
Vx=V0cosa0
Vy=V0sina0-gt



The Attempt at a Solution


I figured part A and the answer is 1.5 m, but for Part B I got -0.93 m/s when I attempted, but the actual answer is -0.89 m/s. I used Vy=V0sina0-gt, where t is .66 s which you get from doing Part A. Help!
 
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Rounding t to .66 is inadequate. If you wish to have 2 digit accuracy in the final answer you need at least 3 digits in calculations.

Oh, I get -0.895 m/s which I would round to .90 so I disagree with the -.89, too!
 

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