Question

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?

Answer 1

Answer:

Explanation:

Given

acceleration of rocket(a)$=5 m/s^2$

At h=1000 m rocket burn out

$v^2-u^2=2as$

$v^2-0=2\times 5\times 1000$

$v=\sqrt{10^4}=100 m/s$

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

v=u+at

$100=0+5\times t$

$t=\frac{100}{5}=20 s$

(c)Rocket maximum altitude

$v^2-u^2=2as$

here u=100 m/s

v=0

$a=9.8 m/s^2$

$s=\frac{v^2}{2g}$

$s=\frac{100^2}{2\times 9.8}$

[tex]s=510.204

Therefore maximum altitude=510.204+1000=1510.204 m