Question

A coin is thrown in a vertically upward direction with a velocity of $5{\text{m s}}^{-1}$. If the acceleration of the coin during its motion is $10{\text{m s}}^{-2}$ in the downward direction, what will be the height attained by the coin and how much time will it take to reach there respectively?

• A. $0.25\text{m},1.5\text{s}$
• B. $1.25\text{m},0.5\text{s}$
• C. $1.25\text{m},0.75\text{s}$
• D. $1.5\text{m},0.25\text{s}$

Answer 1

The correct option is B $1.25\text{m},0.5\text{s}$

Here,
Initial velocity, $u=5{\text{m s}}^{-1},$ and
acceleration, $a=-10{\text{m s}}^{-2}$
At the highest point,
final velocity, $v=0$
Using, $v^2=u^2+2as,$
$0=5^2+[2\times (-10)\times s]$
$\Rightarrow 20s=25$
$\Rightarrow$ Distance, $s=\frac{25}{20}=1.25m$
Hence, the height attained by the coin is $1.25\text{m}$.
Let, the time taken be $t$.
using $v=u+at$,
$0=5-10t$
$\Rightarrow t=\frac{5}{10}=0.5\text{s}$
Thus, time taken by the coin to reach there is $0.5\text{s}$.