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A rocket accelerates upward from rest at a rate of 5 m/s^2. At a height of 1000m, the engines burn out and it coasts the rest of the way with an acceleration of -9.8 m/s^2. a. What is the rocket's velocity at burnout (h=1000m) in meters/second? b. For how many seconds did the rocket engine burn? c. What is the rocket's maximum altitude?

Respuesta :

Answer:

Explanation:

Given

acceleration of rocket(a)[tex]=5 m/s^2[/tex]

At h=1000 m rocket burn out

[tex]v^2-u^2=2as[/tex]

[tex]v^2-0=2\times 5\times 1000[/tex]

[tex]v=\sqrt{10^4}=100 m/s[/tex]

(b) time to reach v=100 m/s

v=u+at

[tex]100=0+5\times t[/tex]

[tex]t=\frac{100}{5}=20 s[/tex]

(c)Rocket maximum altitude

[tex]v^2-u^2=2as[/tex]

here u=100 m/s

v=0

[tex]a=9.8 m/s^2[/tex]

[tex]s=\frac{v^2}{2g}[/tex]

[tex]s=\frac{100^2}{2\times 9.8}[/tex]

[tex]s=510.204

Therefore maximum altitude=510.204+1000=1510.204 m

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