Answer:
(a) t = 2.97s
(b) h = 43.3 m
Explanation:
Let t be the time it takes to fall a distance h, then t - 1 (s) is the time it takes to fall a distance of h - 0.56h = 0.44 h
For the ball to fall from rest a distance of h after time t
Also for the ball to fall from rest a distance of 0.44h after time (t-1)
We can substitute the 1st equation into the 2nd one
and divide both sides by g/2
t = 2.97 or t = 0.6
Since t can only be > 1 s we will pick t = 2.97 s
(b) 
F = applied force on the wheelbarrow in forward direction = 600 N
f = frictional force pushing in backward direction of wheelbarrow = ?
m = mass of the wheelbarrow = 28 kg
Since wheelbarrow moves forward at a constant velocity
a = acceleration of the wheelbarrow in forward direction = 0 m/s²
force equation for the motion of wheelbarrow in forward direction is given as
F - f = ma
inserting the above values in the formula
600 - f = (28) (0)
600 - f = 0
f = 600 N
hence the frictional force comes out to be 600 N
U need to write better u egg head kid or tell your mom
F = ko *Q_1 * Q_2 / d^2
Q_2 = F*d^2 / [ko*Q_1] = 7.2 N * (0.2m)^2 / [9.0 x 10^9 N/m^2C^2 * 4.0 x 10^-6C] = 8x10^-6C = 8 microC
