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An object is dropped from the top of the tower of height 80 m. One second later, another object is thrown downwards with some velocity. The two objects reach the ground simultaneously. The velocity with which the second object was thrown will be (take $g = 10 m / s^{2}$ )

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11.67 m / s

8.9 m / s

5.67 m / s

10.8 m / s

answer is A.

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Detailed Solution

The velocity with which the second object was thrown will be

11.67 m / s

.
Assuming that for the first particle there is no velocity

u = 0

, for the first particle.
So, from third law of motion ,

s = ut + \frac{1}{2} g t^{2}

\Rightarrow s = 0 \times t + \frac{1}{2} g t^{2}

\Rightarrow s = \frac{1}{2} g t^{2}

…

(1)

Here

s =

displacement of the first particle which is

80 m .

t =

time taken and

g =

gravitational acceleration which is

10 m / s^{2}

here.
So, time taken by first object
By equation

(1)

t = \sqrt{\frac{2 s}{g}}

…..

(2)

Put the value of

t

s

and

g

in equation

(2)

, we get,

t =

\sqrt{\frac{2 (80)}{10}}

t = 4 s

Let the initial velocity of the second particle is . After

4 - 1 = 3 s

It travelled

80 m

s = ut + \frac{1}{2} g t^{2}

….

(3)

After putting values of

t

s

and

g

in equation

(3)

, we get

\Rightarrow 80 = u (3) + \frac{1}{2} (10) (3)^{2}

\Rightarrow 3 u = 80 - 45

\Rightarrow 3 u = 35

\Rightarrow u = 11.67 m / s

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