Y^2 = Yo^2 + 2g*h = 0
Yo^2 - 19.6*23.2 = 0
Yo^2 = 454.72
Yo = 21.3 m/s = Ver. component of initial velocity.
Yo = Vo*sin A = 21.3 m/s
48*sin A = 21.3
sin A = 0.44,425
A = 26.4o = Angle at which golf was hit.
Vo = 48m/s[26.4o]
Xo = 48*Cos26.4 = 43.0 m/s.
V = Xo + Yi = 43 + 0i = 43 m/s at max. ht.
a. KE = 0.5M*V^2 = 0.5*0.047*43^2 = 43.5 J.
b. Y^2 = Yo^2 + 2g*h = 0 + 19.6(23.2-8) = 298
Y = 17.3 m/s = Ver. component.
V = sqrt (Xo^2 + Y^2)
Xo = 43 m/s
Y = 17.3 m/s
Solve for V.
A 47.0 g golf ball is driven from the tee with an initial speed of 48.0 m/s and rises to a height of 23.2 m.
(a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.
(b) What is its speed when it is 8.0 m below its highest point?
2 answers
v=51.51m/s
1 answer