A ball is thrown straight upward and returns to the thrower's hand after 3.20 s in the air. A second ball is t

Question

thrown at an angle of 29.0° 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, solve for starting v of ball one. Since it will reach apogee halfway through the flight
0=v0-g*3.2/2
v0=g*3.2/2

Now the vertical v of the second ball must be equal to the v0 of the first
V*sin(29)=g*3.2/2

j