You must have formed that equation somehow.
If h(t) is the displacement of the man as a function of time, then h'(t) is his velocity and h''(t) his acceleration, agreed?
We are told his acceleration is -5ft/s^2 --> h''(t) = -5
Integrating with respect to (wrt) time gives velocity: h'(t) = -5t + C
(where C is an arbitrary constant).
Following?
We are told his initial velocity, that is, his velocity at t=0, is 10ft/s^2.
So substitute t=0 into h'(t) to solve for C:
-5(0) + C = 10 --> C = 10
And we now have h'(t) = -5t + 10
Setting h'(t)=0 -- because his velocity will be zero at the top of his jump -- gives the time it takes for him to reach the top. You did this and got t=2.
Integrating h'(t) gives us h(t).
The integral wrt t of -5t is -(5/2)t^2.
The integral wrt t of 10 is 10t.
And we need to introduce another arbitrary constant, D.
h(t) = -(5/2)t^2 + 10t + D
His initial height is 0 (from wherever his displacement is measured from, presumably the ground).
Substitute into h(t) to solve for D:
0 = -(5/2)(0) + 10(0) + D --> D = 0
(Are you seeing a pattern with these initial conditions being used to find the arbitrary constant?)
And we have a constant-free equation for h(t):
h(t) = -(5/2)t^2 + 10t + 0
\(\displaystyle \mbox{ }\,\,\)= -(5/2)t^2 + 10t
See how you were practically there?
(Although this looks really long, soon you're do it almost instantly)
Finally, we said at t=2 he reaches his max height. So substituting t=2 into h(t) will give us that height.
h(2) = -(5/2)(2^2) + 10(2)
\(\displaystyle \mbox{ }\,\)= 10
(Note that we could have just factorised the quadratic for h(t) and found the vertex of the parabola from there, but why not use calculus.)
So his maximum height is 10ft.
Just because this looks complete does not mean you're not allowed to ask.
