I have issues with my physics homework. Details below

label Physics
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A ball is projected horizontally from the edge of a table that is 1.00 m high, and it strikes the floor at a point 1.20 m from the base of the table.
a.) What is the initial speed of the ball?
b.) How high is the ball above the floor when its velocity vector makes a 45.0 degree angle with the horizontal?
(Please help asap! And also, please include a formula and explanation. Thanks!!)

Oct 9th, 2015

Thank you for the opportunity to help you with your question!

First you calculate how long it takes the ball to fall 1 meter starting from zero vertical speed

The formula to do this is given by; (d=½at²)

And note that the ball travels 1.21m horizontally in time. Speed is determined by diving distance by time

thus; s=1/2at^2+vit 

a. 1=1/2(9.8)t^2+0(t)

t=0.452s

1.2=1/2(0)t^2+vi(0.425)
vi=2.66m/s

b. We have the horizontal speed i.e. 2.66m/s

What we need to calculate is the vertical speed. We use the following formula

vi^2=vf^2+2as

0=2.66^2+9.8.... s

s=0.72m

but the table is I meter

thus 1-0.72

=0.28m


Please let me know if you need any clarification. I'm always happy to answer your questions.
Oct 9th, 2015

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