Astone dropped from the height of 90 m with velocity does it reach to the ground

Explanation and

To solve this problem, we use the kinematics equation:

v^2-u^2=2aS  (1)

where

assuming the positive direction is downwards)

v = final velocity (when it reaches ground)

u = 0 = initial velocity (when it was dropped, or released, with zero vertical velocity).

a = 9.81 m/s^2 = acceleration due to gravity (on earth's surface)

S = 90m = distance travelled.

Substitute values into (1)

v^2 - 0^2 = 2*(9.81)*90

v^2 = 1765.8 m^2/s^2

v = sqrt(1765.8) m/s

= 42.02 m/s

= 42 m/s  (to two significant figures).

c) up is the direction the magnetic force is acting on the charge.