Agolfer 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?

The equation for the maximun height is given by the parametric equation of parabola

Step-by-step explanation:

We have that

with

getting the general equation

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

so if the tangent of theta is less than that fraction the ball won´t clear the wall.

Finally for those data we have:}

x= 1

step-by-step explanation: convert 5\frac{11}{12}5

​12

​

​11

​​   to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

​c

​

​b

​​ =

​c

​

​ac+b

​​

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{5\times 12+11}{12}3.25+0.5x+1

​3

​

​1

​​ +

​6

​​5

​​ x=

​12

​5×12+11

​​ 2 simplify 5\times 125×12 to 6060

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{60+11}{12}3.25+0.5x+1

​3

​​1

​​ +

​6

​5

​​ x=

​12

​​60+11

​​3 simplify 60+1160+11 to 7171

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{71}{12}3.25+0.5x+1

​3

​​1

​​ +

​6

​​5

​​ x=

​12

​​71

​​4 simplify \frac{5}{6}x

​6

​​5

​​ x to \frac{5x}{6}

​6

​5x

​​ 3.25+0.5x+1\frac{1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+1

​3

​

​1

​​ +

​6

​5x

​​ =

​12

​​71

​​ 5 convert 1\frac{1}{3}1

​3

​​1

​​   to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

​c

​​b

​​ =

​c

​​ac+b

​​3.25+0.5x+\frac{1\times 3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

​3

​​1×3+1

​​ +

​6

​​5x

​​ =

​12

​71

​​ 6 simplify 1\times 31×3 to 33

3.25+0.5x+\frac{3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

​3

​​3+1

​​ +

​6

​​5x

​​ =

​12

​71

​​ 7 simplify 3+13+1 to 44

3.25+0.5x+\frac{4}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

​3

​​4

​​ +

​6

​5x

​​ =

​12

​71

​​8 simplify 3.25+0.5x+\frac{4}{3}+\frac{5x}{6}3.25+0.5x+

​3

​​4

​​ +

​6

​​5x

​​   to \frac{13.75}{3}+\frac{4x}{3}

​3

​​13.75

​​ +

​3

​4x

​​ \frac{13.75}{3}+\frac{4x}{3}=\frac{71}{12}

​3

​​13.75

​​ +

​3

​​4x

​​ =

​12

​71

​​ 9 simplify \frac{13.75}{3}

​3

​​13.75

​​   to 4.5833334.583333

4.583333+\frac{4x}{3}=\frac{71}{12}4.583333+

​3

​​4x

​​ =

​12

​71

​​ 10 subtract 4.5833334.583333 from both sides

\frac{4x}{3}=\frac{71}{12}-4.583333

​3

​

​4x

​​ =

​12

​

​71

​​ −4.583333

11 simplify \frac{71}{12}-4.583333

​12

​

​71

​​ −4.583333 to \frac{4}{3}

​3

​4

​\frac{4x}{3}=\frac{4}{3}

​3

​​4x

​​ =

​3

​4

​​ 12 multiply both sides by 33

4x=\frac{4}{3}\times 34x=

​3

​4

​​ ×3

13 simplify \frac{4}{3}\times 3

​3

​​4

​​ ×3 to \frac{12}{3}

​3

​​12

​​ 4x=\frac{12}{3}4x=

​3

​​12

​​ 14 simplify \frac{12}{3}

​3

​​12

​​   to 44

4x=44x=4

15 divide both sides by 44

x=1

​​