Answer:
the maximum height ymax reached by the rocket is 176.4 meters.
Explanation:
The maximum height ymax reached by the rocket can be calculated using the following formula:
ymax = (acceleration * time^2) / 2
Substituting the given values, we get:
ymax = (58.8 m/s^2 * 6.00 s^2) / 2
= 176.4 m
Answer:
Approximately [tex]7.41 \times 10^{3}\; {\rm m}[/tex].
Explanation:
If initial velocity, final velocity, and time taken are all found, then the change in height (displacement) can be found as:
[tex]\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}[/tex].
For example, during the [tex]t = 6.00\; {\rm s}[/tex] of constant acceleration at [tex]a = 58.8\; {\rm m\cdot s^{-2}}[/tex], initial velocity was [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] and final velocity would be:
[tex]\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}[/tex].
Displacement during this period of time would be:
[tex]\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\times 10^{3}\; {\rm m}\end{aligned}[/tex].
During the next part of the flight, initial velocity was [tex]352.8\; {\rm m\cdot s^{-1}}[/tex] (from the first part of flight) and final velocity would be [tex]0\; {\rm m\cdot s^{-1}}[/tex] when the rocket reaches maximum height. Acceleration was given to be [tex]a = (-9.80\; {\rm m\cdot s^{-2}})[/tex](negative since the rocket is accelerating downward,) but time is not known. Apply the following equation to find the change in height (displacement):
[tex]\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\times 10^{3}\; {\rm m}\end{aligned}[/tex].
The total change in height would be approximately [tex](1.058\times 10^{3}\; {\rm m}) + (6.350\times 10^{3}\; {\rm m}) \approx 7.41\times 10^{3}\; {\rm m}[/tex].
