A ball is projected horizontally from the edge of a table thatis 0.45m high , an

Question

A ball is projected horizontally from the edge of a table thatis 0.45m high , and it strikes the floor at apoint 0.79 from the base of the table. The acceleration ofgravity is 9.8m/s2. What is the initial speed of theball? Answer in units of m/s. Part 2 How high is the ball above the floor when its velocityvector makes a -35.0164 degree angle with the horizontal? answer inunits of m. A ball is projected horizontally from the edge of a table thatis 0.45m high , and it strikes the floor at apoint 0.79 from the base of the table. The acceleration ofgravity is 9.8m/s2. What is the initial speed of theball? Answer in units of m/s. Part 2 How high is the ball above the floor when its velocityvector makes a -35.0164 degree angle with the horizontal? answer inunits of m.

Explanation / Answer

Height (H) = 0.45m Distance is (S) = 0.79m The acceleration due to gravity (g) =9.8m/s2 The initial speed of the ball is                           v2 -u2 = 2as Here v =0                                   u2 = 2gs                                    u = 2*9.8*0.79                                        = 3.934m/s b) When the ball makes angle with the horizontal then                                 h = u2sin2/2g                                    = (3.934)2sin2(-35.0164)/2(9.8)                                    = 0.2599m                                    ˜ 0.260m

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