A rocket is shot straight up from the earth with a net acceleration (= acceleration by the rocket engine - gravitational pullback) of 7m/sec during the initial stage of flight until the engine cut out at t = 10 sec. How high will it go, air resistance neglected?

599.7 m

approximately 600 m

Explanation:

initial speed of the rocket = 0

net acceleration upwards = 7 m/s^2

the engine cuts out 10 sec after take off

maximum height reached = ?

we neglect air resistance

To get the velocity of the rocket at the point where the engine cuts off, we use the equation

v = u + at

where

v is the velocity at this point where the engine stops = ?

u is the initial velocity of the rocket from rest = 0 m/s

a is the net acceleration upwards = 7 m/s^2

t is the time the engine runs = 10 s

substituting, we have

v = 0 + (7 x 10)

v = 70 m/s

to get the distance from the ground to this point, we use the equation

= + 2as

where

v is the final velocity at the the height where the engine is cut out = 70 m/s

u is the initial speed at the ground = 0 m/s

a is the net acceleration on the rocket = 7 m/s^2

s is the distance from the ground to this point

substituting, we have

= + 2(7 x s)

4900 = 14s

s = 4900/14 = 350 m

After this point when the engine cuts out, the rocket experiences an acceleration proportional to the acceleration due to gravity 9.81 m/s^2 downwards, and slows down gradually before coming to a stop at the maximum height.

To get this height, we use the equation

= - 2gs   (the negative sign is due to the downward direction of the acceleration g)

where

v is the final velocity at the maximum height = 0 m/s (it comes to a stop)

u is the speed at the instance that the engine is cut out = 70 m/s

g is the acceleration due to gravity = 9.81 m/s^2

s is the distance from this point to the maximum height

substituting values, we have

= - 2(9.81 x s)

0 = 4900 - 19.62s

4900 = 19.62s

s = 4900/19.62 = 249.7 m

The maximum height that will be reached = 350 m + 249.7 m = 599.7 m

approximately 600 m