Acertain freely falling object, released from rest, requires 1.35 s to travel the last 38.5 m before it hits the ground.
(a) find the velocity of the object when it is 38.5 m above the ground. (indicate the direction with the sign of your answer. let the positive direction be upward.)

m/s

(b) find the total distance the object travels during the fall.

Given: s = displacement = 38.5m
t = 1.35s
Vo = ?
a = 9.8 m/s^2

a) Using the following formula for free-fall motion:

s = Vo*t + (1/2)a*t^2
38.5 = Vo * 1.35 + (1/2)(9.8)*1.35^2
Vo = -21.9 m/s

b)  Solving for the distance prior to reaching 38.5m before reaching the ground:

Vf = Vo + at

Where:
Vf = 21.9 m/s
Vo = 0
a = 9.8
t = ?

21.9 = 0 + (9.8)*t
t = 2.24s

Solving for s:

s = (Vo + Vf)/2 *t
s = (0 + 21.9)/2 * 2.24
s = 24.528 m

Solving for the total displacement:

Total displacement = 24.528 m + 38.5 m
Total displacement = 63.028 m

Caused conformity rates to go down. apex ; )

Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge