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

Answer 1

Answer: (D)

1

Answer 2

Time taken by the particle to reach a height h from the surface of Earth is obtained from $h=ut-\frac{1}{2}gt^{2}$
Here, u = 9 m/s, h = 4 m, g = 10 m/s
$\therefore\quad4=9t-\frac{1}{2}\times10\times t^{2}$
4 = 9t - 5t or 5t - 9t + 4 = 0
5t - 5t - 4t + 4 = 0
$\Rightarrow\quad$ 5t(t - 1) - 4(t - 1) = 0
$\therefore\quad$ 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.