Question

A particle is thrown vertically up with an initial velocity $9 \,m/s$ from the surface of Earth (take $g = \,10 \,m/s^2$).The time taken by the particle to reach a height of $4\, m$ from the surface second time (in seconds) is

• A. 1.3
• B. 1.2
• C. 1.1
• D. 1.0

Answer 1

Answer: (D)

Time taken by the particle to reach a height $h$ from the surface of Earth is obtained from $h=u t-\frac{1}{2} g t^{2}$
Here, $u=9\, m / s , h =4\, m , g =10 \,m / s ^{2}$
$\therefore 4=9 t-\frac{1}{2} \times 10 \times t^{2} $
$4=9 t-5 t^{2} $ or $ 5 t^{2}-9 t+4=0$
$5 t^{2}-5 t-4 t+4=0 $
$\Rightarrow 5 t(t-1)-4(t-1)=0 $
$\therefore t=1 \,s$ or $ t=\frac{4}{5} s$
Hence, the time taken by the particle to reach a height of $4 \,m$ from the surface second time is $1 \,s$.