Question

A stone is thrown straight downward with initial speed 8.0m/s from height of 20m find (a) the time it takes to reach the ground and (b) speed with which it strikes.

Answer 1

Force = - m g = -m * 9.81 m/s^2 = m a
so
a = -9.81 m/s^2
v = Vi + a t = -8.0 - 9.81 t
h = Hi + Vi t + (1/2) a t^2 = 20 - 8.0 t - 4.9 t^2
ground when h = 0
4.9 t^2 + 8.0 t - 20 = 0
t = 1.36 seconds is positive time root of quadratic
v = -8.0 - 9.81 (1.36) = -21.3 meters/seconds
speed is absolute value of velocity = 21.3 m/s
ALTERNATELY
energy at top = m g Hi + (1/2) m Vi^2 = m(9.81*20+.5*64)
= 228 * m Joules
energy at bottom = (1/2) m v^2
so
m v^2 = 2 * 228 * m
v^2 = 456
v = 21.3 m/s (again)
average speed = .5 (8+21.3) = 14.7 m/s
distance = 20
so time = 1.36 seconds (again)