Question #799aa

2 Answers
Jun 19, 2015

Answer:

I found #44.7m#
But check my maths.

Explanation:

Ok, I considered that the absolute acceleration of the rockets is #12 m/s^2# and used (with #i=# initial and #f=# final):
#y_f-y_i=v_it+1/2at^2#
#20=0t+1/2*12t^2#
#t=1.82s#
Now I use #v_f=v_i+at#
#v_f=0+12*1.82=22m/s#

at this point the acceleration stops and the rockets reaches the maximum height (under influence of gravity #g#) where #v_f=0#; I use again: #v_f=v_i+at# but now
#v_i=22m/s#,
#g=-9.8 m/s^2#
and #v_f=0#:
So:
#0=22-9.8t#
#t=2.24s#
I use this time #t# into #y_f-y_i=v_it+1/2at^2# again:
#y_f-20=22*2.24-1/2*9.8(2.24)^2#
#y_f=44.7m#
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Jun 19, 2015

Answer:

The maximum height of the rocket is 44.5 m.

Explanation:

Here's how you can solve this problem without using the time of accelerated flight and the time of free fall.

You know that the rocket starts at ground level with an initial velocity equal to zero. After 20 m of accelerated flight, the rocket reaches a height of 20 m.

This means that you can determine the speed of the rocket when it shuts down its engines by

#v_1^2 = underbrace(v_0^2)_(color(blue)("=0")) + 2 * a * h_1#, where

#h_1# - the distance it travelled having the initial acceleration.

After it shuts down the engines, the rocket continues its upward flight under the influence of gravity, which means that the gravitational acceleration, #g#, will act upon the rocket.

At maximum height, the speed of the rocket will be equal to zero. This means that you can write

#underbrace(v_2^2)_(color(blue)("=0")) = v_1^2 - 2 * g * h_2#, where

#h_2# - the distance it covered after shutting down the engines;

Since you the expression for #v_1^2#, you can determine #h_2# by

#0 = underbrace(2 * a * h_1)_(color(blue)("="v_1^2)) - 2 * g * h_2#

Therefore,

#h_2 = (cancel(2) * a * h_1)/(cancel(2) * g) = (12cancel("m"/"s"^2) * "20 m")/(9.8cancel("m"/"s"^2)) = "24.5 m"#

The maximum height the rocket reached was

#H = h_1 + h_2 = 20 + 24.5 = color(green)("44.5 m")#