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The discus thrower throws the discus at a velocity with a magnitude 12.2 m/s at a height 5.15 ft above the ground. In terms of meters, 5.15 ft is equal to 1.57 m. As the discus moves there is an acceleration of 9.8 m/s^2 acting on it in a direction vertically downwards. It is assumed that air resistance is negligible.
Let the angle at which the discus is thrown to the horizontal be A. The horizontal component of the velocity is 12.2*cos A and the vertical component is 12.2*sin A. As the discus rises the vertical component of its velocity decreases, it is equal to 0 at the highest point of the trajectory and becomes negative as the discus moves lower from that point. Let the maximum height be H and the time taken to reach it be T. Use the equation V = U + a*T, V = 0, U = 12.2*sin A and a = -9.8
T = `(12.2*sin A)/9.8` .
At the highest point, the height of the discus is `(12.2*sin A)*((12.2*sin A)/9.8) - (1/2)*9.8*((12.2*sin A)/9.8)^2`
The time taken by the discus to touch the ground as it falls is `sqrt((2*(12.2*sin A*((12.2*sin A)/9.8) - (1/2)*9.8*((12.2*sin A)/9.8)^2 + 1.57))/9.8)`
=> `sqrt(((12.2*sin A)^2/9.8 + 3.14)/9.8)`
The total time of travel of the discus is `(12.2*sin A)/9.8 + sqrt(((12.2*sin A)^2/9.8 + 3.14)/9.8)`. The horizontal distance traveled by the discus or its range is
R = `((12.2*sin A)/9.8 + sqrt(((12.2*sin A)^2/9.8 + 3.14)/9.8))*12.2*cos A`
=> `((12.2*sin A)/9.8 + sqrt((12.2*sin A)^2 + 30.772)/9.8)*12.2*cos A`
=> `(12.2*sin A + sqrt((12.2*sin A)^2 + 30.772))*(61/49)*cos A`
To maximize R, the equation `(dR)/(dA) = 0` has to be solved.
This is very difficult to manually. Instead, MS Excel has been used to determine the angle A for which R is maximum, the result is 43.7 degrees.
It should be noticed that this is very close to the angle 45 degrees which would have been the angle at which the discus should have been thrown from ground level. As it is thrown from a height of 5.15 ft, the angle is reduced by around 1.3 degrees.
The discus should be thrown at an angle of 43.7 degrees to the horizontal to maximize the range.
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