Given a rocket with constant acceleration after t = 4, when will it hit the ground? A rocket is launched straight upward. During the first four seconds of powered flight, its height is given by:


*

*$h(t) = 16.1t^2 − 1.75t^3$ 

*The function is valid when $0 ≤ t ≤ 4$

*$t$ in seconds and $h$ in feet.


At the instant when $t = 4$ seconds, the fuel cuts off. From that point in time onward, the rocket has constant acceleration of $-32.2 ft/s^2$.
When does it hit the ground?
Given this information I have hypothesized:


*

*$h'(t) = -32.2t$

*$h''(t) = -32.2$

*At the instant fuel cuts off $h(4)=145.6$


Taking two anti-derivatives, the $h(t)$ for $t > 4$ is now: 


*

*$h(t)= -16.1t^2+403.2$


Solving for $h(t)=0$ at $t ≥ 4$ gives $t=5.004$ but this is not the correct answer. Can someone help me understand what I'm missing?
Graph of $h(t)$ and $y=0$ intersection provided: 
https://www.desmos.com/calculator/hddvv1de1z
 A: Hint: What is the velocity at $t=4$?
A: For $t \ge 4$:
$$
\dot{v} = a_4 \Rightarrow \\
\int\limits_{4\text{s}}^t \dot{v} \, d\tau 
=  v - v_4 
= \int\limits_{4\text{s}}^t a_4 \, d\tau \Rightarrow \\
v = a_4 (t - 4\text{s}) + v_4 \Rightarrow \\
h = \frac{a_4}{2}(t-4\text{s})^2 + v_4 (t-4\text{s}) + h_4 \\
$$
where $a_4 = a(4\text{s})=−32.2\text{ft}/\text{s}^2$, $v_4 = v(4\text{s})=44.8\text{ft}/\text{s}$, $h_4 = h(4\text{s})=145.6\text{ft}$.
We then need to solve
$$
0 = \frac{a_4}{2}(t-4\text{s})^2 + v_4 (t-4\text{s}) + h_4
$$
for $t$ which means
$$
0 = \left( (t-4\text{s})+\frac{v_4}{a_4}\right)^2
+ \frac{2h_4}{a_4}- \left(\frac{v_4}{a_4}\right)^2 \iff \\
t = 4\text{s} + \frac{\pm \sqrt{v_4^2-2 h_4 a_4}- v_4}{a_4} 
$$
where we pick the negative root to get $t = 8.7 \text{s} \ge 4 \text{s}$. (Note: $a_4 < 0$)
A: $h′(t)=−32.2t$ - that's for t>4. For t<=4:
$h′(t)=32.2t - 5.25t^2$
so, $h(4) = 145.6$ 
$v(4) = h′(4) = 44.8$
let x be the time from the moment of cutoff.
$145.6+(44.8-32.2x)x = 0, x>0$
Positive root of this equation is ~2.933.
So, the answer will be $x+4 = 6.933$
