Question

A rocket is fired vertically up from the ground with a resultant vertical acceleration of $10m/{s^2}$ . The fuel is finished in $1\min$ and it continues to move up.
(a) What is the maximum height reached?
(b) After how much time from then will the maximum height be reached?
( Take $g = 10m/{s^2}$ )
A. $36km,1\min$
B. $6km,1\min$
C. $30km,1\sec$
D. $36km,1\sec$

Answer 1

Hint This is a great question as per standards. First we have to use the second equation of motion to find total distance travelled with the fuel. Then after that we have to apply the third equation of motion to find the distance after the fuel gets over. Then we have to add then=m to find the total distance. After that we have to use the first equation of motion to find the time.

Complete step-by-step solution :Firstly we have to sort out what we have got :
Resultant vertical acceleration of rocket : $10m/{s^2}$
Fuel is finished within time : $1\min$
: The total distance travelled by the rocket within the first $1\min$ :
The acceleration is $10m/{s^2}$
The time in seconds $= 60\sec$
{h_1} = \dfrac{1}{2}a{t^2} \\\ {h_1} = \dfrac{1}{2} \times 10 \times {60^2} \\\ {h_1} = 18000m \\\ {h_1} = 18km \\\
Now, the velocity after the fuel is over would be :
$v = at \\\ v = 10 \times 60 \\\ v = 600m/s \\\$
After the fuel will get over than acceleration will get in the opposite direction:
$a = - 10m/{s^2}$
Now we just have to put the values in the equation for $v = 0m/s$ as at the peak point it would be so :

$${v^2} = {u^2} + 2a{h_2} \\\ 0 = {600^2} + 2 \times 10 \times {h_2} \\\ 360000 = 20{h_2} \\\ {h_2} = 18000m \\\$$

Now the total distance will be equal to :
$= {h_1} + {h_2} \\\ = 18000m + 18000m \\\ = 36km \\\$
So the total distance would be $36km$
: We already0 know that the final velocity would be zero, therefore to find the time travelled in that period we just have to use the first equation of motion :
$v = u - gt \\\ 0 = 600 - 10 \times t \\\ 10t = 600 \\\ t = \dfrac{{600}}{{10}} \\\ t = 60s \\\ t = 1\min \\\$
Therefore the total time taken : $t = 1\min$
Hence, the correct option would be A.

Note:- Here we have to take care that the given acceleration is already the resultant acceleration, the one after the calculation of gravity. Apart from that we should consider ideal situations until no exception is given.