Question

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

• A. $s=1.5m,t=0.25s$
• B. $s=0.25m,t=1.5s$
• C. $s=1.25m,t=0.75s$
• D. $s=1.25m,t=0.5s$

Answer 1

The correct option is D $s=1.25m,t=0.5s$

Given, initial velocity, $u=5m{s}^{-1}$, acceleration, $a=-10m{s}^{-2}$.

Let '$v$' be the final velocity and '$s$' be the height attained.
At the highest point of motion, $v=0$

Using third equation of motion, $v^2=u^2+2as,$
$0=5^2+[2\times (-10)\times s]$
$\Rightarrow s=1.25m$
Hence the height attained by the coin is $1.25m$.

Let the time taken be $t$.
Using the first equation of motion,
$v=u+at$
$0=5-10t$
$\Rightarrow t=0.5s$
$\therefore$ Time taken by the coin is $0.5s$.