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

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

$v=0+2\times 6=12 m/s$

After this object will start moving under gravity

height reached in first 6 s

$s=ut+\frac{at^2}{2}$

$s=0+\frac{2\times 6^2}{2}$

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

$a=9.8 m/s^2$

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

$s_0=\frac{144}{2\times 9.8}=7.34 m$

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