Question

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

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

Answer 1

The correct option is D

S = 1.25 m, t = 0.5 s

Here,

$u=5m/s,a=-10m/s^2$

At the highest point, $v=0$

Using $v^2=u^2+2as,$

$0=5^2+[2\times (-10)\times s]$

$\Rightarrow s=1.25m$

Hence the height attained by the coin is $1.25m$.

Let time taken be $t$.

$v=u+at$

$0=5-10t$

$t=0.5s$

Time Taken by the coin to reach there is $0.5s$.