A particle is thrown vertically up with an initial velocity 9 m/s from the surface of Earth (take g = 10 m/s2).The time taken by the particle to reach a height of 4 m from the surface second time (in seconds) is

Question

A particle is thrown vertically up with an initial velocity 9 m/s from the surface of Earth (take g = 10 m/s2).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

Tardigrade Admin · Accepted Answer

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}

∴ 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

∴ 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

.

Q. A particle is thrown vertically up with an initial velocity 9m/s from the surface of Earth (take g=10m/s2).The time taken by the particle to reach a height of 4m from the surface second time (in seconds) is

1.3

1.2

1.1

1.0

Q. 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