what will be the additional missle height?

Question

A missle being launched reaches a vertical velocity of 100m/s when the rocket engine shuts off. To what additional height will the missle go before it stops going up and falls back to Earth?

Answer 1

Initial velocity = 100 m/s
This is no different than shooting a projectile straight up with an initial velocity of 100 m/s.

s = v^2 / 2g
s = 100^2 / 19.6
s = 10000 / 19.6
s = 510.2 m

Answer 2

a good question in deed. it helops tou understand that when the engines stop due to inertia the rocket still moves but its weight poses a drug and eventually brings it in a stop.
there are various way of solvnig this problem.
i prefere the energetic approach.

i declare the point where the engine shuts off as the zero plane reference level for potential energy and then apply the law of the conservation of kinetice energy.

taking into consideration a frictionless enviroment - that the weight remains the same and a perfect and ideal wolrd we come to the conslusion that.

1/2 m * U^2 = m*g*h

where (be carefull) the H is the extra distance travles since point zero for this case is taken the point where the engine shuts off.
now the mass cancels out and we come to the point where

h = u^2 / 2 g (usefull equation in order to find height)

now the g is a problem since 100 m /s must be in great heights but as i told you before let us consider it as an ideal world and take it as 9.81 m / s^2


therefore

h = 100*100 / 2* 9.81 = 509.68m

Answer 3

Just use the same numbers as you would to figure how long it would take gravity to accelerate something to 100m/s if you dropped it off a cliff, and multiply that times the average speed of 50m/s.
Stopping a rising object is the same as accelerating a dropped object in the unreal perfect world of physics problems.