Question

g A rock is launched from the edge of a cliff with an initial speed of 64 m/s at an angle of 60° above the horizontal, as illustrated. What is most nearly the rock’s displacement from its initial position after 6.0 s?

Answer 1

s= 544 m ∅=39°

Step-by-step explanation:

Initial velocity= Vi = 64 m/s

projection angle=∅ =60°

Horizontal velocity= Vx =Vicos∅

Vx= 64Cos(60°)

Vx=32 m/s

Horizontal distance covered in 6 sec:

Sx= Vx*t (horizontal velocity remain constant)

Sx= 32*6 =192 m

Vertical Velocity= Vy = Vi*Sin∅

Vy= 64 * Sin(60°)

Vy=55.43 m/s

Vertical distance covered:

Sy= Vy * t +1/2 a*t^2 (here a=-g)

Sy= 55.43*6 -1/2(9.8*6^2)

Sy= 156.2 m

Total displacement from starting point is:

s=
`\sqrt{(Sx)^(2)+(Sy)^(2)  }`

s=
`\sqrt{(192)^(2)+(156.2)^(2)  }`

s= 247.5 m ∠39°