A 47.0-g golf ball is driven from the tee with an initial speed of 47.0 m/s and rises to a height of 21.1 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 6.85 m below its highest point?
here,
mass of golf ball , m1 = 47 g = 0.047 kg
initial speed of ball, u = 47 m/s
h = 21.1 m
a)
using conservation of energy
the kinetic energy at highest point , Ke = initial kinetic energy - final potential energy
Ke = 0.5 * m * u^2 - m * g * h
Ke = 0.5 * 0.047 * 47^2 - 0.047 * 9.81 * 21.1 J
Ke = 42.2 J
b)
let the speed at h1 = 6.85 m be v
using conservation of energy
initial kinetic energy = final kinetic energy + final potential energy
0.5 * m * u^2 = 0.5 * m * v^2 + m * g * h1
47^2 - 2 * 9.81 * 6.85 = v^2
solving for v
v = 45.5 m/s
the speed of ball is 45.5 m/s
Get Answers For Free
Most questions answered within 1 hours.