Arock thrown straight up takes 4.2 s to reach its maximum height what was its initial velocity

Consider the upward direction of motion as positive and downward direction of motion as negative.

a = acceleration due to gravity in downward direction = - 9.8

v₀ = initial velocity of rock in upward direction = ?

v = final velocity of rock at the highest point = 0

t = time to reach the maximum height = 4.2 sec

Using the kinematics equation

v = v₀ + a t

inserting the values

0 = v₀ + (- 9.8) (4.2)

v₀ = 41.2

well clustered together.

points close on a scatter diagram

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j