A golfer hits his ball with speed v o at an angle 6 above the horizontal ground. Assuming that the angle 0 is fixed and that air resistance can be neglected, what is the minimum speed v o (min) for which the ball will clear a wall of height h, a distanced away? Your solution should get into trouble if the angle 0 is such that tan 0 < h 1 d. Explain. What is v o (min) if 6 = 25°, d = 50 m, and h = 2 m?

Answer:The equation for the maximun height is given by the parametric equation of parabola[tex]h=\frac{v_{0}^2\sin^2(\theta)}{2g}[/tex]Step-by-step explanation:We have that[tex]h=v_{0}t_{h}\sin(\theta)-\frac{1}{2}gt{h}^2[/tex]with [tex]t_{h}=\frac{v_{0}\sin(\theta)}{g}[/tex]getting the general equation[tex]h=\frac{v_{0}^2\sin^2(\theta)}{2g}[/tex]For sure if the tangent of theta is less than h/d the ball doesn´t pass through the wall, this because we have the following trigonometric relation[tex]\tan(\theta)=\frac{h}{d}[/tex]so if the tangent of theta is less than that fraction the ball won´t clear the wall.Finally for those data we have:}[tex]v_{0}=\frac{2}{\sin(25)}=4.73[/tex]