Question

a rocket is fired vertically from the ground with a restaurant vertical acceleration of 10 ms-2 the fuel is finished in 1minute and it continues to move up.what is maximum height reached?​

Answer 1

Answer :-

18000m

Explanation :-

Given :

Initial velocity = 0m/s

Acceleration = 10m/s

Time = 1min = 60s

Acceleration = 10m/s^2

To Find :

Maximum height,h = ?

Solution :

According to the first equation of motion,

$\boxed{\sf{}v=u+at}$

where,

v is the final velocity,

u is the initial velocity,

a is the acceleration and,

t is the time taken.

Put their values and find “v”

$\sf{}\implies v=0+10\times 60$

$\sf{}\implies v=600m/s$

Now we know the final velocity so let’s find out the maximum height reached by the rocket.

According to the second equation of motion,

$\boxed{\sf{}v^2=u^2+2as}$

where,

v is the final velocity,

u us the initial  velocity,

a is the acceleration and,

s is the distance covered.  

Put their values and find “h”

$\sf{}\implies (600)^2=0+2\times 10\times h$          [here s is same as h]

$\sf{}\implies 360000=20h$

$\sf{}\implies \dfrac{360000}{20}= h$

$\sf{}\implies h =18000m$

Therefore,maximum height reached by the rocket is 18000m or 18km