- #1

brotherbobby

- 668

- 158

- Homework Statement
- Bond and Scaramanga get into a helicopter which begins to accelerate upward at ##a\;\text{m/s}^2## for ##3\;\text{s}## at which point Scaramanga gets thrown out. He bites the dust ##4.92\;\text{s}## after liftoff. (i) What was ##a##? (ii) From what height did Scaramanga get ejected?

- Relevant Equations
- The kinematic equations for uniform acceleration ##a_0## where all terms have their usual meanings. Just note that ##x(t_0)=x_0## and that ##v(t_0)=v_0##. We can choose ##t_0=0## here.

\begin{align}

&v(t)=v_0+a_0(t-t_0)\\

&x=x_0+v_0(t-t_0)+\dfrac{1}{2}a_0(t-t_0)^2\\

&v^2(x)=v^2_0+2a_0(x-x_0)\\

\end{align}

**S**) gets thrown out with a speed ##v_0## at time ##t_1=3\;\text{s}## from the helicopter which is accelerating at ##a\;\text{m/s}^2## starting from rest from the ground. After a time of ##t_2=1.92\;\text{s}## from then,

**S**reaches the ground. Using equation ##(2)## above, we can write $$-h=v_0t_2-\frac{1}{2}gt_2^2\qquad\text{(4)}$$where using ##(1)##, ##v_0=at_1## and using ##(2)## again ##h=\frac{1}{2}at_1^2##.

Substituting in ##(4)##,

\begin{align*}

-\frac{1}{2}at_1^2 &= at_1t_2-\frac{1}{2}gt_2^2\\

\Rightarrow \frac{1}{2}gt_2^2 &= at_1t_2+\frac{1}{2}at_1^2\\

\Rightarrow a(t_1t_2+\frac{1}{2}t_1^2) &=\frac{1}{2}gt_2^2\\

\Rightarrow at_1(2t_2+t_1)&=gt_2^2\\

\Rightarrow a &= \dfrac{gt_2^2}{t_1(2t_2+t_1)}\\

\Rightarrow a &= \dfrac{10\times 1.92^2}{3(2\times 1.92+3)}\\

\Rightarrow &\boxed{a=2.54\;\text{m/s}^2}

\end{align*}

The height at which the throwing off took place : ##h = \frac{1}{2}at_1^2=\frac{1}{2}\times 2.54\times 3^2=\boxed{11.43\;\text{m}}##

**Doubt :**The problems don't match with those in the text that I copy and paste to the right.

**Request :**I'd like to know where have I gone wrong. Was it in the meaning of the term "liftoff"?