Assuming the simple model where the object is projected with an initial velocity of u metres/second at an angle of x to the horizontal, and that the only force acting on it after that is gravitational acceleration, g = 9.81 metres/second^2, then h = [u*sin(x)]^2/(2*g) metres.
If the launch is vertical then x = pi/2 radians and h = u^2/(2*g) metres.
A projectile reaches its maximum height when the vertical velocity , Vy , is zero. Basic kinematics
gives :
ymax = ( Vy0 )^2 / ( 2 ) ( g )
where Vy0 is the initial vertical velocity and g is the gravitational acceleration.
45 degrees.
The horizontal component of a projectile's velocity doesn't change, until the projectile hits somethingor falls to the ground.The vertical component of a projectile's velocity becomes [9.8 meters per second downward] greatereach second. At the maximum height of its trajectory, the projectile's velocity is zero. That's the pointwhere the velocity transitions from upward to downward.
The value of the vertical speed at the highest point of the projectile's trajectory is the lowest speed at the maximum height reached.
15.42 degrees
You can't unless you know gravity and air pressure as well.
If a projectile takes 8 seconds to reach its maximum height, it will take another 8 seconds to return to its original elevation. Presuming it is lauched from flat ground and returns to the ground, its total time in flight would be 16 seconds. If it is launched from a hill, or at a hill, more information would be needed.
The max height depends only on the angle and speed at release. It doesn't depend on the projectile's weight.
Without special road permits, 13'6" is maximum.
30"
It depends. If the projectile goes straight up and straight down, its velocity will be zero at the top. If the projectile is a baseball about halfway between the pitcher and the bat, its velocity might be 150 km/h.
C = 2pi * r
You cannot without knowing what the shape is.