Question

A Rocket is launched to travel vertically upwards with a constant velocity of 20 m/s. After travelling for 35 seconds rocket develops a snag and its fuel supply is cut off. The rocket then travels like a free body. What is the total height achieved by it? After what time of its launch will it come back to the earth? ( g= 10 m/s²⁾

Answer 1

We know, S = ut + 1/2 at²

This means the height = 20 * 35 + 0 (since rocket move upward with constant velocity)

h = 700m

After this ,

h = ut - 1/2 gt2 ( because the body goes up)

We know , v² - u² = -2gh

0² - 20² = -2 * 10 * h (here, final velocity means the velocity at the top that is taken as 0...... intial velocity is 20 because of the power given by fuel..... now it experiences no external force other than gravity........ height here is the distance travelled after the fuel suppl is cut off)

-400 = -20h

h = 20m

Thus, total height = 700 + 20 = 720m (first answer)

Now we need to calculate the time taken to reach the top when fuel supply is cut off.....

We know , v = u -gt

this means 0 = 20 - 10 * t

0 = 20 - 10t

10t = 20

t = 2secs

Thus total time while going up from ground = 35 + 2 = 37secs

Now we need to calculate the time taken for rocket to reach bottom from top,

h = ut + 1/2 gt²

720 = 0*t + 1/2 * 10 * t * t

720 = 0 + 5t²

720/5 = t²

144 =t²

t =12

Thus the total time = 37 secs (total time for up) + 12 secs(total time for down)

that is 49 secs (second answer)