Answer:
a) 
b) 
c) 
Explanation:
From the exercise we know the initial velocity of the projectile and its initial height

To find what time does it take to reach maximum height we need to find how high will it go
b) We can calculate its initial height using the following formula
Knowing that its velocity is zero at its maximum height



So, the projectile goes 1024 ft high
a) From the equation of height we calculate how long does it take to reach maximum point



Solving the quadratic equation



So, the projectile reach maximum point at t=2s
c) We can calculate the final velocity by using the following formula:


Since the projectile is going down the velocity at the instant it reaches the ground is:

Can something have energy even if it's not moving?
All moving objects have kinetic energy. When an object is in motion, it changes its position by moving in a direction: up, down, forward, or backward. ... Potential energy is stored energy. Even when an object is sitting still, it has energy stored inside that can be turned into kinetic energy (motion).
Does a book at rest have energy?
A World Civilization book at rest on the top shelf of a locker possesses mechanical energy due to its vertical position above the ground (gravitational potential energy).
Does a book lying on a table have energy?
The book lying on a desk has potential energy; the book falling off a desk has kinetic energy.
Answer:
5.09 x 10⁵ Nm²/C
Explanation:
The electric flux φ through a planar area is defined as the electric field Ε times the component of the area Α perpendicular to the field. i.e
φ = E A
From the question;
E = (8.0j + 2.0k) ✕ 10³ N/C
r = radius of the circular area = 9.0m
A = area of a circle = π r² [Take π = 3.142]
A = 3.142 x 9² = 254.502m²
Now, since the area lies in the x-y plane, only the z-component of the electric field is responsible for the electric flux through the circular area.
Therefore;
φ = (2.0) x 10³ x 254.502
φ = 5.09 x 10⁵ Nm²/C
The electric flux is 5.09 x 10⁵ Nm²/C