Question

An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Determine the speed at this time and the maximum height at which it reaches.

Answer 1

Answer:43.34 m

Step-by-step explanation:

Given

acceleration(a)
`=2 m/s^2`

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

`v=0+2* 6=12 m/s`

After this object will start moving under gravity

height reached in first 6 s

`s=ut+(at^2)/(2)`

`s=0+(2* 6^2)/(2)`

s=36 m

After fuel run out distance traveled in upward direction is

`v^2-u^2=2as_0`

here v=0

u=12 m/s

`a=9.8 m/s^2`

`0-12^2=2(-9.8)(s)`

`s_0=(144)/(2* 9.8)=7.34 m`

`s+s_0=36+7.34=43.34 m`