A rocket is launched from the earth surface so that it has an acceleration of 19.6m/s square.if it's engine is switched off after 5 seconds of its launch,then the maximum height attained by the rocket will be...
Plz solve it to be the brainliest
AbhishGS:
what is the ans at last given?
ok let me try
rough
the main difficulty was to get which formula to be used
Answers
Answered by
13
accln= 19.6 m/s time t =5sec
Hmax = u square /2g
for initial velocity.
a=u/t i.e. 19.6=u/5
:. u=19.6×5=98m/s
:. Hmax=98 square/2(10)
9604/20=480.2m
Hmax = u square /2g
for initial velocity.
a=u/t i.e. 19.6=u/5
:. u=19.6×5=98m/s
:. Hmax=98 square/2(10)
9604/20=480.2m
then nearest value will ur ans
For MCQs we should not get real ans but nearest to it
good luck for studies
Answered by
21
Initially
a1= 19.6
t1 = 5sec
u1= 0
v1 = u1 + a1t1
v1= 19.6*5 = 98
S1 = ut + at^2/2
S1 = 19.6*25/2
Now after the engine stops
a = g= -9.8
u = v1=98
v = 0 after this point rocket will start falling
v = u + at
t = -98/-9.8 = 10sec
S2 = ut + at^2/2
S2 = 98*10 - 9.8*100/2
S2 = 980 - 9.8*50
S2 = 980 - 19.6*25
now
max height = total distance travelled
= S1 + S2
max height = 980 m
a1= 19.6
t1 = 5sec
u1= 0
v1 = u1 + a1t1
v1= 19.6*5 = 98
S1 = ut + at^2/2
S1 = 19.6*25/2
Now after the engine stops
a = g= -9.8
u = v1=98
v = 0 after this point rocket will start falling
v = u + at
t = -98/-9.8 = 10sec
S2 = ut + at^2/2
S2 = 98*10 - 9.8*100/2
S2 = 980 - 9.8*50
S2 = 980 - 19.6*25
now
max height = total distance travelled
= S1 + S2
max height = 980 m
i think AbishGS will agree now
ok
i am not as good in phy as maths
just tried
cbse students can solve easily
they know the concepts. more probable than hsc students
this answer is wrong, the height will be 735 m. you forgot to divide S1 by two in the last lines. hence, max height is S1 +S2 = 245 + 480 = 735m
^you are right , well spotted
tnx guys
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