Answer to Question #252391 in Civil and Environmental Engineering for Joma

Question

A golf ball is driven horizontally from an elevated tee with a speed of 25 m/s. It strikes the fairway 2.5 sec later.

(a) How far has it fallen vertically?

(b) How far has it travelled horizontally?

(c) Find the horizontal and vertical components of its velocity just before it strikes the fairway. 

Answer 1

$y = ut+ \frac12at²\\ y = 0(t) + \frac12(9.8)+(2.5)²\\ y = 0+ 4.9(6.25)\\ y = 30.625m$

(A)

Vertical distance will be : 30.625 m

(B)

Horizontal distance will be 62.5 m

(C) Components

Distance 

$d = \sqrt{x^2 + y^2 + z^2}$

Components of given forces 

$From \space \dfrac{F_x}{x} = \dfrac{F_y}{y} = \dfrac{F_z}{z} = \dfrac{F}{d}$

$F_x = \dfrac{x \, F_x}{d}$

Resultant

$R = \sqrt{{R_x}^2 + {R_y}^2 + {R_z}^2}$

Direction cosines of the resultant

$\cos \theta_x = \dfrac{R_x}{R} = -0.394$

$\cos \theta_y = \dfrac{R_y}{R} = 0.762$

$\cos \theta_z = \dfrac{R_z}{R} = -0.514$