physics
posted by Nat on .
I need to calculate the horizontal, vertical and resultant velocity of a ball thrown a distance of 8.23m in a time of 1.73sec (assuming the ball lands at the same height from which it is thrown). The angle of the throw is unknown.
Please help!

u = 8.23/1.73 = 4.76 m/s horizontal speed
in air for 1.73 s
so going up for .865 s
v = Vi  g t
0 = Vi  9.81(.865)
so
Vi = 8.49 m/s initial speed up
for resultant initial speed
s^2 = u^2 + Vi^2
= 22.66 + 72.1 = 94.7
so
s = 9.73 m/s
T = angle to horizontal
sin T = 8.49/9.73
T = 60.8 degrees above horizontal