Ejected out from an accelerating spaceship

[brotherbobby]

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}

Let me copy and paste the problem as it appeared in the text to the right.

I start by drawing the diagram of the problem. Scaramanga (drawn as 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"?

 

[erobz]

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}

View attachment 339718Let me copy and paste the problem as it appeared in the text to the right.




View attachment 339719I start by drawing the diagram of the problem. Scaramanga (drawn as 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 : $h = \frac{1}{2}at_1^2=\frac{1}{2}\times 2.54\times 3^2=\boxed{11.43\;\text{m}}$

View attachment 339720Doubt : 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"?

Your interpretation seems correct to me.

 

[TSny]

@brotherbobby : The answers given in the text appear to come from assuming that S has zero velocity relative to the ground just after being thrown out of the helicopter. We don't have information about how S was thrown out. But, your assumption that S has zero velocity relative to the helicopter seems more reasonable. I agree with your answers.

 

[PeroK]

Is this yet another question concocted by someone who doesn't actually understand physics?

 

[kuruman]

Is this yet another question concocted by someone who doesn't actually understand physics?

The flaw in the supposed answer has been discussed here. After solving this problem, I agree with EulerJr's numerical answer of 1.76 m/s² in the reddit.com link and disagree with @brotherbobby's answer of 2.54 m/s².

 

[brotherbobby]

The flaw in the supposed answer has been discussed here. After solving this problem, I agree with EulerJr's numerical answer of 1.76 m/s² in the reddit.com link and disagree with @brotherbobby's answer of 2.54 m/s².

Yes, am sorry, I made calculation errors. The acceleration of the spaceship is $a = 1.76\;\rm{m/s^2}$ and the height of ejection is $\dfrac{1}{2}\times 1.76\times 3^2 = \rm{7.92\; m}$.

The crucial point of the exercise was the author's mistake. He assumed that an object thrown out of a moving vehicle is at rest relative to ground and not at rest relative to the vehicle.