A test rocket is fired straight up from rest. The net acceleration is 20 m/s2 upward and continues for 4.0 seconds, at which time the rocket engines cease firing. What maximum elevation does the rocket reach?

Respuesta :

aachen

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