Asked by Nat
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!
Please help!
Answers
Answered by
Damon
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
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
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