High School Physics
posted by Crying Girl on .
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 =-(
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.
thank you so much!!! i feel so stupid...but at least you helped. Your awesome thanks for saving my day.
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
that answer is completely wrong.
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
you messed up in the finishing part when 21.5 should be added with 1.83 then moved over