A 50.7-g golf ball is driven from the tee with an initial speed of 51.2 m/s and rises to a height of 21.8 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 6.55 m below its highest point?

Answer 1

a. Y^2 = Yo^2 + 2g*h = 0 @ max. ht.
Yo^2 = -2g*h = 19.6 * 21.8 = 427.28
Yo = 20.7 m/s. = Ver. component of initial velocity.

Yo = 51.2*sin A = 20.7 m/s.
sinA = 20.7/51.2 = 0.40373
A = 23.8o.

Vo = 51.2[23.8o]
Xo = 51.2*cos23.8 = 46.8 m/s. = Hor.
component of initial velocity.

V = Xo + Yi
V = 46.8 + 0i @ max. Ht.
V = Xo = 46.8 m/s @ max. Ht.

KE=0.5m*V^2=0.5*0.0507*(46.8)^2=55.5 J.

b. Y^2 = Yo^2 + 2g*h
Y^2 = (20.7)^2 + (-19.6*(21.8-6.55)
Y^2 = 428.49 - 298.9 = 129.59
Y = 11.38 m/s. = Ver. component of velocity.

V = Xo + Yi
V = 46.8 + 11.4i = 48.2 m/s[13.7o].

Answer 2

Correction:

b. Y^2 = Yo^2 + 2g*h
Y^2 = 0 + 19.8(21.8-6.55) = 299
Y = 17.3 m/s. = Ver. component.

V = Xo+Yi = 46.8 + 17.3i
V = sqrt(46.8^2+17.3^2) = 49.9 m/s.