math problem....can you help me?

Question

If a rock is thrown upward with an initial velocity of 32 feet per second from the top of a 40-foot building, write the height equation using this information.

Answer 1

H=S+Ho ,if the time is t ,gravity acc. g , Vo initial vel.,then
H=( Vo .t -1/2 . g t^2)+ 40 as it is projected upward
H= (32.t -1/2 .32 t^2) +40= 23t -16t^2 +40 ft.

Answer 2

You do not state the angle at which the rock is thrown. Is it thrown straight up so that when it lands it will land on the building? Isit thrown horizontally so that it will land somewhere on the ground presumably at a point that is at the same level as the base of the building? Or is it thrown at an angle somewhere between horizontal (zero degrees) and vertical (90 degrees).

Let z = the angle at which the rock is thrown.
Then its position at any time t is given by:
x = Vocos z)t and y = -1/2g t^2 + (Vo sin z)t + 40,
where Vo = initial velocity and g= gravitational constant.

You can eliminate the parameter t and get:
y = -gx^2/(2Vo cos^2z) + x tan z +40 which is the equation of a parabola.

Answer 3

The equation of motion is:
s = s(initial) + v(initial)t + (1/2)at^2
Your equation becomes:
s = 40 + 32t + (1/2)gt^2
Where g is the acceleration of gravity.