Physics free fall

Question

Hello everyone , I have solved this question, and would like someone to discuss with me whether my answer is right,

the question says ,
A rocket ascends vertically from rest at ground level with acceleration 10 m.s-2 ,Calculate :
a- the height of the rocket after 4 s .

b- The speed of the rocket after 4 s

After the rocket has been moving for 4 s a component breaks off and falls to the ground .

c- Calculate the further time which elapses before the component hits the ground .


My answers,

So first we need to calculate the height of the rocket after 4 s , so we already have
t = 4
a = 9.8 m.s-2 '' As it moving downwards so it is positive ''
u = 0 '' from rest ''

so we can say :

S = 1/2at2

then we get S = 78 m .

for b , we have ,

S = 78 m
u = 0
V = ?
a = 9.8
t= 4
so ,
v2 = u2 + 2as

v2 = 0 + 2(9.8)(78)
v2 = 1528.8
v = 39 m.s-1

for c ,

I found that there are some few aspects of understanding this question ,
First I thought after the rocket ascend in 4 s there is component that falls down, therefore the component fell from the same distance that the rocket arrived after 4 s , If it makes sense to say that we can calculate the distance till the rocket arrive to the ground as we have v= 0 and u=v = 39
so I put
0 = (39)2 +2(9.8)s

but the thing is I'm going to get a negative answer which doesn't make sense at all,


and there is another aspect of understanding this question ,

which is when the ball falls so it falls from the same point where the rocket fell, umm which I find a bit difficult to calculate the whole distance , because what I said is when it arrives the ground its v = 0 , and obviously it started from rest u = 0 , So how could that be ,


Thank you , and hope for someone to help me solve this .

Answer 1

A rocket ascends vertically from rest at ground level with acceleration 10 m/s²

Calculate :
a) the height (vertical displacement) of the rocket after 4 s.

 

The first part looks right, except you are using 9.8m/s² instead of the given acceleration of the Rocket. This might throw off all your calculations. You want the Rockets acceleration as it moves upward not the acceleration of gravity.

 

Vi = 0

a = 10m/s²

yo = 0

 

y(t) = yo + Vi*t + (1/2) * a * t²

 

y(4) = 0 + 0 *(4) + (1/2) * 10 * (4)²

 

y(4) = 5 * 16

y(4) = 80 m

 

b) The speed of the rocket after 4 s

 

V(t) = Vi + a * t

 

V(4) = 0 + 10(4)

V(4) = 40 m/s

 

c) After the rocket has been moving for 4 s a component breaks off and falls to the ground.

Calculate the further time which elapses before the component hits the ground.

Must choose a reference point. I'll choose the breaking point of the rocket part, 80m in the air.

 

Rocket part initial velocity is then;

Vi = 40 m/s

acceleration due to gravity = -9.8 m/s2

 

displacement = -80 m

(final - initial) = (0-80) = -80

 

S(t) = Vi*t + (1/2) * a * t²

 

-80 = 40t + (1/2)(-9.8)t²

 

0 = -4.9t² + 40t + 80

 

Using the quadratic equation, you arrive at only one valid answer (other is negative)

 

t = 9.825 ~ 10 seconds