Question

A rocket takes off vertically from the launchpad with no initial velocity but a constant upward acceleration of 2.25 m/s2. At 15.4 s after blastoff, the engines fail completely so the only force on the rocket from then on is the pull of gravity. (a) What is the maximum height the rocket will reach above the launchpad? (b) How fast is the rocket moving at the instant before it crashes onto the launchpad? (c) How long after engine failure does it take for the rocket to crash onto the launchpad?

Answer 1

Answer:

Explanation:

Given

Initial velocity (u)=0

t=15.4 s

acceleration(a)=$2.25 m/s^2$

(a)maximum height the rocket will reach is s

v=u+at

$v=0+2.25\times 15.4=34.65m/s$

$v^2-u^2=2gs$

$0-34.65^2=-2\times 9.81\times s$

s=61.19 m

(b) velocity of rocket before it crashes on to launchpad

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

Now rocket will be under complete control of gravity

here u=0 as rocket has reach its maximum height

up to 15.4 s rocket has covered a distance of 266.805

$v^2=2\times 9.81\times (61.19+266.805)$

$v=\sqrt{6435.261}=80.22 m/s$

(c)Time taken will be addition of rocket reaching maximum height plus rocket reaching launchpad

$t_1=\frac{34.65}{9.81}=3.53 s$

Now from maximum height to launchpad

$v=u+at_2$

$80.22=0+9.81\times t_2$

$t_2=8.177 s$

t=3.53+8.177=11.71 s