Abasketball referee tosses the ball straight up for the starting tip-off. at what velocity (in m/s) must a basketball player leave the ground to rise 1.32 m above the floor in an attempt to get the ball? m/s â

For this analysis,
g = 9.8 m/s²
Wind resistance is ignored.
All quantities are measured positive upward.

h = 1.32 m, the vertical height

Let u = vertical launch velocity of the basketball player.
At the maximum height of 1.32 m, the vertical velocity is zero.
Therefore
u² - 2gh = 0
u² = 2gh = 2*9.8*1.32 = 25.872
u = 5.0865 m/s

  5.1 m/s (nearest tenth)

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