Homework Help Forum: High School Physics

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

  
  

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