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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