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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?

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jonathanjmc30

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]

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