Ok, I believe i found my source of error. When solving for initial velocity equation (7) should turn out as following:
7) Vi = sqrt( ((g)(xf^2)) / ((2)(cos^2(angle))((-yi) - (xf)(tan(angle))))
I left out the cos^2(angle). I redid the calculations and yielded the answer in the back of the...
So the given data is as follows:
angle = 42 degrees
yf = 0m
yi = 2.1m
xi = 0m
xf = 17m
I used the given data to produce the following six equations where "t" is time and "V" is velocity and "g" is the acceleration due to gravity -9.8m/s^2.
1) Vfx = (Vi)(cos(angle))
2) xf =...
A shotputter projects the shot at 42 degrees to the horizontal from a height of 2.1 m. It lands 17 m away horizontally. Next, he gives it the same initial speed but changes the angle to 40 degrees. What effect does this have on the horizontal range?