Question

I need help with math

Answer 1

We have the following problem:

Using the kinematic equation, we know that for movements with constant acceleration the position is

$h(t)=h_0+v_0t-\frac{gt^2}{2}$

I already wrote the variables convenient for our problem, where

t = time (s)

h₀= initial height (m)

v₀ = initial velocity (m/s)

g = gravity acceleration (m/s²)

h = height after t seconds

We also know that

t = independent variable

h₀= 10m

v₀ = 56 m/s

g ≅ 10 m/s²

h = dependent variable

Therefore our quadratic will be

$\begin{gathered} h(t)=10+56t-\frac{10t^2}{2} \\ \\ h(t)=10+56t-5t^2 \end{gathered}$

Now we can answer the question in fact.

a)

The rocket will hit the ground when the height is equal to zero, hence, h(t) = 0. In fact, we are looking for the zeros of the quadratic

$\begin{gathered} 10+56t-5t^2=0 \\ \\ t=\frac{-56\pm\sqrt{56^2-4\cdot(-5)\cdot(10)}}{2\cdot(-5)} \\ \\ t=\frac{-56\pm\sqrt{3136+200}}{-10} \\ \\ t=\frac{56\pm\sqrt{3336}}{10} \\ \\ t=11.38\text{ s or }-0.18\text{ s} \end{gathered}$

See that we have one negative zero, but we will ignore that because it's physically impossible, therefore the rocker will reach the ground after 11.38 seconds

b)

To find the maximum height we must find the max value of the quadratic, we know that the vertex of a quadratic is its max/min, also, we can write it as

$\begin{gathered} y_V=-\frac{\Delta}{4a} \\ \\ x_V=-\frac{b}{2a} \end{gathered}$

Where

$(x_V,y_V)$

Is the vertex. Here, we want to find the y coordinate, because y here is the height, therefore

$\begin{gathered} \max h=-\frac{\Delta}{4a} \\ \\ \operatorname{\max}h=-\frac{3336}{4\cdot(-5)} \\ \\ \operatorname{\max}h=\frac{3336}{20} \\ \\ \operatorname{\max}h=166.8\text{ m} \end{gathered}$

The max height of the rocket is 166.8m

c)

Now we have a similar problem, it's also the vertex but now the coordinate x, that here, represents the time, then

$\begin{gathered} t_{\max}=\frac{-b}{2a} \\ \\ t_{\operatorname{\max}}=\frac{-56}{2\cdot(-5)} \\ \\ t_{\operatorname{\max}}=\frac{56}{10} \\ \\ t_{\operatorname{\max}}=5.6\text{ s} \end{gathered}$

The rocket will reach the maximum height after 5.6 seconds.