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A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 58.8 m/s^2 . The acceleration period lasts for time 6.00 s until the fuel is exhausted. After that, the rocket is in free fall.

Find the maximum height ymax reached by the rocket. Ignore air resistance and assume a constant free-fall acceleration equal to 9.80 m/s^2 .

User Jimijazz
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1 Answer

4 votes

Answer:

Approximately
7.41 * 10^(3)\; {\rm m}.

Step-by-step explanation:

If initial velocity, final velocity, and time taken are all found, then the change in height (displacement) can be found as:


\begin{aligned}(\text{displacement}) &= \frac{(\text{average velocity})}{(\text{time taken})} \\ &= \frac{(1/2)[(\text{initial velocity}) + (\text{final velocity})]}{(\text{time taken})}\end{aligned}.

For example, during the
t = 6.00\; {\rm s} of constant acceleration at
a = 58.8\; {\rm m\cdot s^(-2)}, initial velocity was
u = 0\; {\rm m\cdot s^(-1)} and final velocity would be:


\begin{aligned}v &= (\text{initial velocity}) + (\text{acceleration}) \, (\text{time}) \\ &= (0\; {\rm m\cdot s^(-1)}) + (58.8\; {\rm m\cdot s^(-2)})\, (6.00\; {\rm s}) \\ &= 352.8\; {\rm m\cdot s^(-1)}\end{aligned}.

Displacement during this period of time would be:


\begin{aligned}(\text{displacement}) &= \frac{(\text{average velocity})}{(\text{time taken})} \\ &= \frac{(1/2)[(\text{initial velocity}) + (\text{final velocity})]}{(\text{time taken})} \\ &= \frac{(1/2)\, (0\; {\rm m\cdot s^(-1)} + 352.8\; {\rm m\cdot s^(-1)})}{(6.00\; {\rm m\cdot s^(-1)})} \\ &\approx 1.058* 10^(3)\; {\rm m}\end{aligned}.

During the next part of the flight, initial velocity was
352.8\; {\rm m\cdot s^(-1)} (from the first part of flight) and final velocity would be
0\; {\rm m\cdot s^(-1)} when the rocket reaches maximum height. Acceleration was given to be
a = (-9.80\; {\rm m\cdot s^(-2)})(negative since the rocket is accelerating downward,) but time is not known. Apply the following equation to find the change in height (displacement):


\begin{aligned}(\text{displacement}) &= \frac{[(\text{final velocity})^(2) - (\text{initial velocity})^(2)]}{2\, (\text{acceleration})} \\ &\approx \frac{(0\; {\rm m\cdot s^(-1)})^(2) - (352.8\; {\rm m\cdot s^(-1)})^(2)}{2\, (-9.80\; {\rm m\cdot s^(-2)})} \\ &\approx 6.350* 10^(3)\; {\rm m}\end{aligned}.

The total change in height would be approximately
(1.058* 10^(3)\; {\rm m}) + (6.350* 10^(3)\; {\rm m}) \approx 7.41* 10^(3)\; {\rm m}.

User Ebelanger
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