Question

Help please...
h(t)=c-(d-4t)^2
At t=0 a ball is thrown in the air from an initial height of 6 feet. Its height in feet , after (t) seconds is given by the function h above, in which c and d are positive constants. If the ball reached its maximum height of 106 feet at t=2.5, what was its height at the time t=1. Will medal.

Answer 1

Well, assuming the ball falls after thrown.. we can say it 'stopped' raising at time 2.5 and started falling.
So we can say it has a 'local maximum' in there, and therefore a derivative of 0.
$$
h(t) = c - (d-4t)^2 \\
h'(t) = \bigg[ (d-4t)^2 \bigg]' = 2(d - 4t) \cdot(-4) = -8d +32t \\
h'(t) = 0 \implies 32t - 8d = 0 \implies 8d = 32t \implies d = 4t
$$
We know we should have \(h'(t) = 0\) at \(t = 2.5\) so:
$$
h(2.5) = 0 \implies d = 4 \cdot 2.5 = 10 \\
h(t) = c - (10 - 4t)^2
$$
We also know we had an initial height of 6 feet which means at time 0:
$$
h(0) = c - (10-0)^2 = c - 10^2 = c - 100 = 6 \implies c = 106 \\
h(t) = 106 - (10 - 4t)^2 \\
h(1) = 106 - (10 - 4 \cdot 1) ^2 = 106 - 6^2 = 106 - 36 = 70
$$

Answer 2

Thank you. I was trying to determine how to solve this using the derivative at 0

Answer 3

The word 'assuming' is somewhat wrong here.. we can see there is a local maximum by how h(t) defined.
Well sure np =)