Posted by Crying Girl on Tuesday, September 15, 2009 at 2:46am.
The men's world record for the shot put, 23.12m , was set by Randy Barnes of the United States on May 20, 1990.
If the shot was launched from 6.00 f above the ground at an initial angle of 42.0 degrees , what was its initial speed?
Someone plz help =(

High School Physics  drwls, Tuesday, September 15, 2009 at 3:05am
There are two unknowns: the initial speed V and the flight time T. You need two equations to solve for them both.
The 6 ft launch height should be comverted to meters (1.83 m)
The horizontal velocity component is Vcos 42 = 0.7431 V
From the length of the shotput record,
0.7431 V T = 23.12 m, so VT = 31.11 m
The vertical component of the initial speed is Vsin42 = 0.6691V
The time T required to hit the ground is given by
0.6691 V*T  4.9 T^2 + 1.83 = 0
I recommend that you substitute 31.11 m for VT in the last equation, and solve for T. Then use that T to compute V.

High School Physics  Crying Girl, Tuesday, September 15, 2009 at 3:13am
thank you so much!!! i feel so stupid...but at least you helped. Your awesome thanks for saving my day.

High School Physics  drwls, Tuesday, September 15, 2009 at 7:12am
You are welcome. Don't cry about physics. It can be fun.
To finish the problem, I got
21.50 4.9 T^2 + 1.83 = 0
4.9 T^2 = 19.67
T = 2.00 s
V = 31.11/2 = 15.56 m/s
Most champion shot putters spin their body while they launch the shot with their arm, to get extra velocity and greater distance

High School Physics  Anonymous, Saturday, September 11, 2010 at 2:26pm
that answer is completely wrong.

High School Physics  mena, Thursday, September 23, 2010 at 7:18am
the answer isn't wrong.it is just rounding errors that give the false answer.if you use all the decimal points, you should get a final answer of 14.5

High School Physics  big pimpin, Monday, September 12, 2011 at 3:39pm
you messed up in the finishing part when 21.5 should be added with 1.83 then moved over
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