A ball is thrown straight upward and returns to the thrower's hand after 2.30 s in the air. A second ball is thrown at an angle of 42.0° with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically?

Answer 1

first ball stops in 2.3/2 = 1.15 seconds
0 = Vi - g t
Vi = 9.81 (1.15) = 11.3 m/s

second ball has same Vi
Vi = V sin 42
V = 11.3/sin 32 = 21.3 m/s

Answer 2

vf=vi+at, 0=vi+)(9,81)(1,15)

Answer 3

damon how you go from sin 42 to sin 32