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A world class shotputter can put a 7.26 kg shot a distance of 22 m. Assume that the shot is constantly accelerated over a distance of 2 m at an angle of 45 degrees and is released at a height of 2 m above the ground. Estimate the weight that this athlete can lift with one hand.

Also, determine the initial angle [tex]\theta[/tex] of the trajectory to maximize the distance R of the put.

So I reasoned the key to the first part is to find [tex]v_0[/tex] so that I can use it to find the force with which the shotput was thrown. I then divide that force by g to find the mass that the shotputter can lift.

I started by finding the total time the shotput was airborne. I solved the following equation for t quadratically:

[tex]y = -\frac{1}{2}gt^2 + v_0 sin\theta t + y_0 = 0[/tex]

I found [tex]t = \frac{v_0 sin\theta + \sqrt{v^2_0 sin^2\theta + 4}}{g}[/tex]. Plugging this value into the equation [tex]x = v_0 cos\theta t = 22[/tex] I chugged through and found [tex]v_0 = 14.6 m/s[/tex].

Now I go back to the first part of the throw, before the shotput is released. Using [tex]v^2 - v_0^2 = 2 a (x - x_0)[/tex] I can arrive at:

Weight shotputter can lift with one arm = [tex] \frac{F}{g} = \frac{ma}{g} = \frac{mv^2}{2gx}[/tex] and I find the mass to be 39.4 kg. The book lists 42.0 kg as the answer so I'm not sure where the discrepancy lies since I didn't round until the very end when I did the problem.

The second part is beyond me. If I take the range equation I used above, namely,

[tex]x = v_0 cos\theta t = \frac{v_0^2}{2g} sin2\theta + \frac{\sqrt{v^2_0 sin^2\theta + 4}}{g}[/tex]

the derivative is horrendous and I'm not sure I can solve it for [tex]\theta[/tex]. Also, the little formula the back of the book has indicates that there must be some other, more tractable, range formula to differentiate. I can't seem to figure out where to start in finding an alternate way of doing it. This also makes me think that there might have been a cleaner way of doing the first part. Am I right in thinking that or am I just missing something obvious here?