A ball is projected horizontally from the edge of a table that is 1.14 m high, and it strikes the floor at a point 1.40 m from the base of the table. It's initial velocity is 2.9 m/s. How high is the ball above the floor when its velocity vector makes a 43.6o angle with the horizontal?

Answer 1

Xo = 2.9 m/s.

Tan 43.6 = Y/Xo = Y/2.9.
Y = 2.9*Tan 43.6 = 2.76 m/s.

V^2 = Vo^2 + 2g*d = 2.76^2.
0 + 19.6d = 7.63, d = 0.389 m.

h = 1.14 - 0.389 = 0.751 m. Above
the floor.