Answer:
[tex]b.)= (\frac{25}{2} , \ -25\frac{\sqrt{3} }{2})\ mph[/tex]
Step-by-step explanation:
Given;
velocity of the diver, v = 25 mph
direction of his dive, θ = 60°
The vertical component of the velocity is given by;
[tex]-V_y = vsin\theta \ \ (below \ horizontal \ is \ in \ negative\ y-direction)\\\\V_y =-(25)(sin 60)\\\\V_y =-(25)(\frac{\sqrt{3} }{2} )[/tex]
The horizontal component of the velocity is given by;
[tex]V_x =vcos\theta\\\\V_x =(25)(cos 60 )\\\\V_x =(25)(\frac{1 }{2} )\\\\V_x = \frac{25}{2}[/tex]
Therefore, the component form of the velocity vector is given by;
[tex](V_x, \ V_y) = (\frac{25}{2} , \ -25\frac{\sqrt{3} }{2})\ mph[/tex]
correct option = [tex]b.)= (\frac{25}{2} , \ -25\frac{\sqrt{3} }{2})\ mph[/tex]
