Arocket-launching vehicle is moving forward at a constant velocity of 5 m/s. a cannon on the vehicle shoots a shell straight up with a velocity of 20 m/s. the shell moves without friction, (no air resistance). how high does the shell go, in m?

For any object thrown upwards where only the force of gravity is acting upon it, uses the following formula for the maximum height attained.

H= v²/2g, where g = 9.81 m/s²

There are two information of velocities are given. However, we use the 20 m/s information because this is the launch velocity. Hence, the solution is as follows:

H = (20 m/s)²/2(9.81 m/s²)
H = 20.4 m

Ibelieve that the correct answer is: c. along plate boundaries

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