Question

From the top of a tower, a stone is thrown up. It reaches the ground in 5 sec. A second stone is thrown down with the same speed and reaches the ground in 1 sec. A third stone is released from rest and reaches the ground in how many seconds????
a)3 sec
b) 2 sec
c) √5 sec
d)2.5 sec​

Answer 1

Answer:

A third stone is released from rest and reaches the ground in $\sf \sqrt{5}$ seconds.

Explanation:

Given:

From the top of a tower, a stone is thrown up. It reaches the ground in 5 sec. A second stone is thrown down with the same speed and reaches the ground in 1 sec.

To Find:

A third stone is released from rest and reaches the ground in how many seconds?

Formula to be used:

$\implies \sf t = \sqrt{t_1 \times t_2}$

Solution:

Given,

• $\sf t_1 = 5 sec$

• $\sf t_2 = 1 sec$

Using the formula,

$\implies \sf t = \sqrt{t_1 \times t_2}$

$\implies \sf t = \sqrt{5 \times 1}$

$\implies \sf t = \sqrt{5} sec$

$\therefore$ A third stone is released from rest and reaches the ground in ${\underline{\bf \sqrt{5}}}$ seconds.

Answer 2

Answer:

A third stone is released from rest and reaches the ground in $\sf \sqrt{5}$ seconds.

Explanation:

Given:

From the top of a tower, a stone is thrown up. It reaches the ground in 5 sec. A second stone is thrown down with the same speed and reaches the ground in 1 sec.

To Find:

A third stone is released from rest and reaches the ground in how many seconds?

Formula to be used:

$\implies \sf t = \sqrt{t_1 \times t_2}$

Solution:

Given,

• $\sf t_1 = 5 sec$

• $\sf t_2 = 1 sec$

Using the formula,

$\implies \sf t = \sqrt{t_1 \times t_2}$

$\implies \sf t = \sqrt{5 \times 1}$

$\implies \sf t = \sqrt{5} sec$

$\therefore$ A third stone is released from rest and reaches the ground in ${\underline{\bf \sqrt{5}}}$ seconds.