Question

An elevator which is moving with an upward acceleration of "a $m/s^2$”. A person drops a coin to measure the acceleration 'a $m/s^2$'. Find 'a' if the coin took 0.5s to hit the ground and the coin was dropped from a height of 3 m.

• A. 10 m/s²
• B. 14 m/s²
• C. 5 m/s²
• D. 0 m/s²

Answer 1

The correct option is B

14 m/s²

$U_{\text{lift}}=+U$
$U_{\text{coin}}=+U\left(same\right)$

So $u_{\text{coin/lift}}=U-U=0$
So with respect to lift coin's initial velocity=0
Now $a_{\text{lift}}=+a$
$a_{\text{coin}}=-g$
$a_{\text{coin/lift}}=-g-a$

Now w.r.t. i.e., if I am the lift and I have eyes I will see the coin falling a fall of 3m. So the displacement will be -3 as in negative y-axis. Now, $s=ut+\frac{1}{2}a{t}^2$
$-3=0+\frac{1}{2}(-g-a)\left(\frac{1}{2}{)}^2\right(since,t=0.5sec)$

$\Rightarrow +3=\frac{(g+a)}{g}$
g+a=24
a=24-g=14 $m/s^2$
So lift is going 14 $m/s^2$ in upward direction.