A 46.9-g golf ball is driven from the tee with an initial speed of 52.4 m/s and rises to a height of 28.5 m?

Question

(a) Neglect air resistance and determine the kinetic energy of the ball at its highest poin. (b) What is the speed when it is 5.06 m below its highest point?

Answer 1

a) v^2-u^2=2as
v^2-52.4^2=2*(-9.8)*28.5
v^2= 2745.8-558.6
=2187.2
v=46.76m/s

Kinetic Energy=1/2mv^2
=1/2*46.9*46.76^2
=51,290 Joules

b)height=28.5-5.06=23.44m

v^2-u^2=2as

v^2-52.4^2=2*(-9.8)*23.44
v^2=2745.8-459.42
=2286.4
v=47.8m/s

Answer 2

(a)

When the ball reaches its highest point, its velocity is zero (v = 0 m/s). Since the velocity is zero, so is the kinetic energy (KE = (1/2)mv²). Therefore, KE = 0 J.

(b)

28.5 m - 5.06 m = 23.44 m

v² = vo² + 2aΔy

v² = 52.4² - 2(9.80)(23.44)

v² = (2.75 x 10^3) - 2(9.80)(23.44)

v² = (2.29 x 10^3)

v = 47.9 m/s