What Went Wrong in Solving the Ball Projection Problem?

[Sealy]

Homework Statement

A ball is projected horizontally from the edge of a table that is 0.443 m high, and it strikes the floor at a point 1.84 m from the base of the table.

The acceleration of gravity is 9.8 m/s^2

Homework Equations

a) What is the initial speed of the ball? Answer in units of m/s

b) How high is the ball above the floor when its velocity vector makes a -18.9486 degree angle with the horizontal? Answer in units of m.

The Attempt at a Solution

a) i got it right, and the answer is 6.13

b) this is what i have trouble on. I started by finding Vy using (6.13)(tan -18.9486)=Vy and i got: Vy=-2.10 I then used the formula Vf = Vo + a*t to get time. I got t=-.21 s
Next i used d = Vot + 0.5a*t^2 to solve for distance and i got d= .22

Finally i did 0.443-0.22 and i got 0.22. The answer is not .22.

I need help please. What did i do wrong?

 

[tiny-tim]

Welcome to PF!

Hi Sealy! Welcome to PF!

b) this is what i have trouble on. I started by finding Vy using (6.13)(tan -18.9486)=Vy and i got: Vy=-2.10 I then used the formula Vf = Vo + a*t to get time. I got t=-.21 s
Next i used d = Vot + 0.5a*t^2 to solve for distance and i got d= .22

Finally i did 0.443-0.22 and i got 0.22. The answer is not .22.

I need help please. What did i do wrong?

(probably a rounding error, but anyway …)

there are three standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations …

why not use v² = u² + 2as ?

 

[Sealy]

I tried to redo the problem without rounding until the end and i still got 0.22. I even tried 0.21 and 0.23. I sent a copy of my work and my solution to my teacher today so I'm waiting for her response.

I really think i used the right method.