Posted by **Michaela** on Sunday, November 20, 2011 at 12:39pm.

A ball is projected horizontally from the edge of a table that is 1.15 m high, and it strikes the floor at a point 1.70 m from the base of the table.

(a) What is the initial speed of the ball?

m/s

(b) How high is the ball above the floor when its velocity vector makes a 45.0° angle with the horizontal?

m

- physics -
**Henry**, Monday, November 21, 2011 at 3:36pm
h = Vo*t + 0.5g*t^2 = 1.15m,

0 + 4.9t^2 = 1.15,

t^2 = 0.235,

t = 0.484s = Time in flight.

a. Dh = Xo * t = 1.70m, = Hor distance.

0.484Xo = 1.70,

Xo = 3.5m/s. = Initial hor velocity.

b. tan45 = Y/Xo,

Y = Xo*tan45 = 3.5 * 1 = 3.5m/s = ver.

comp. of velocity.

d = (Vf^2 - Vo^2) / 2g.

d = ((3.5)^2 - 0) / 19.8 = 0.625m.

h = 1.15 - 0.625 = 0.525m. above the floor.

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