Answer:
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
