Given a rocket with constant acceleration after t = 4, when will it hit the ground?

Question

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

Answer 1

Hint: What is the velocity at $t=4$?

Answer 2

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: $a_4 < 0$)

Answer 3

$h′(t)=−32.2t$ - 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 $x+4 = 6.933$