Given that the acceleration due to gravity is 32.2 feet per second per second. Find theheight above the ground of an object thrown upward from a pointS0meters abovethe ground with an initial velocity ofv0meters per second. Hint: Acceleration due togravity is in the downward direction, and therefore will be negative

We have that:

The acceleration is a = 32.2 feet per second or 9.8 m/s^2 (we should use this one because the other numbers are in meters)

S0 is the initial position

V0 is the initial velocity.

We start for the acceleration, we have:

A(t)  -9.8m/s^2

for the velocity, we integrate over time:

V(t) = ( -9.8m/s^2)*t + V0

for the position we integrate again:

P(t) = (1/2)( -9.8m/s^2)*t^2 + V0*t + S0

P(t) is the height above the ground as a function of t, we can find the maximun height by findind at which time the velocity is equal to zero (this means that the object stops going up and starts falling down)

v(t) = 0 = ( -9.8m/s^2)*t + V0

t = V0/( -9.8m/s^2)

Now, we replace this in the position equation and we can find the maximum height,

maxP = (1/2)( -9.8m/s^2)*(V0/( -9.8m/s^2))^2 + V0^2/( -9.8m/s^2) + S0

you can cut your hand off if your not