A 48.7-g golf ball is driven from the tee with an initial speed of 47.5 m/s and rises to a height of 27.5 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 6.60 m below its highest point?
a. Yf^2 = Yo^2 + 2g*h = 0.
Yo^2 = -2g*h = -2*(-9.8)27.5 = 539.
Yo = 23.2 m/s. = Vertical component of
initial velocity.
sin A = Yo/Vo = 23.2/47.5 = 0.48877.
A = 29.3o.
Xo = Vo*Cos A = 47.5*Cos29.3 = 41.4 m/s
= Horizontal component of initial velocity = Total velocity at max ht.
KE = 0.5M*Xo^2 = 0.5*.0487*(41.4)^2 = 41.7 J.
b. V^2 = Vo^2 + 2g*(27.5-6.6).
Vo = 47.5 m/s.
g = -9.8 m/s^2.
V = ?
o/Vo =