Originally Posted by
skeeter no calculus, huh?
since $\displaystyle \Delta y = 0$
$\displaystyle 0 = v_{yo}t - \frac{1}{2}gt^2$
$\displaystyle 0 = t\left(v_{yo} - \frac{1}{2}gt\right)$
$\displaystyle v_{yo} = \frac{1}{2}gt$
$\displaystyle t = \frac{2v_{yo}}{g} = \frac{2v_o\sin{\theta}}{g}$
$\displaystyle \Delta x = v_x \cdot t$
$\displaystyle \Delta x = v_o\cos{\theta} \cdot \frac{2v_o\sin{\theta}}{g}$
$\displaystyle \Delta x = \frac{v_o^2 \cdot 2\sin{\theta}\cos{\theta}}{g}$
$\displaystyle \Delta x = \frac{v_o^2 \sin(2\theta)}{g}$
max value for $\displaystyle \Delta x$ occurs when $\displaystyle \sin(2\theta) = 1$
$\displaystyle \Delta x_{max} = \frac{v_o^2}{g}$
min ball speed will occur when it is at the peak of its trajectory ... when $\displaystyle v_y = 0$ ... min speed would therefore be ... ?
max height occurs at $\displaystyle t = \frac{v_{yo}}{g}$