Respuesta :
Answer:
Maximum elevation, h = 160 meters
Explanation:
Initially, the rocket is at rest, u = 0
Acceleration of the rocket, [tex]a=20\ m/s^2[/tex]
Time, t = 4 s
We need to find the maximum elevation reached by the rocket. It can be calculated using second equation of motion as :
[tex]h=ut+\dfrac{1}{2}\times a\times t^2[/tex]
[tex]h=\dfrac{1}{2}\times 20\times (4)^2[/tex]
h = 160 meters
So, the maximum elevation reached by the rocket is 160 meters. Hence, this is the required solution.
Answer:
486.5 m
Explanation:
Initial velocity is zero as the rocket is fired from rest. u = 0.
Displacement of the rocket during this time:
s = ut +0.5 at²
s = 0+0.5 ×20×4²
s = 160 m
The final velocity at the end of 4 s is:
v = u + at
v = 0 + (20)(4)
v = 80 m/s
This will become the initial velocity for the next half of the motion.
At the maximum elevation, velocity is zero. v = 0
Acceleration due to gravity always acts downwards.
[tex]s=\frac{v^2-u^2}{2a}\\s=\frac{0-80^2}{2\times -9.8} = 326.5 m[/tex]
Thus, the maximum elevation that test rocket would reach is:
326.5 m + 160 m = 486.5 m