Question

a stone is dropped into water from a bridge 44.1 m above the water. Another stone is thrown vertically downward 1 second later .Both strike the water simultaneously, then the initial speed of the second stone is

Answer 1

Hi there!
Here's the answer:


•°•°•°•°•°<><><<><>><><>°•°•°•°•°

We have,

$S = ut + \frac{1}{2}a t^{2}$


Given,
S = 44.1 m
a = g = 9.8 m/s²
and for a freely falling body u =0

Substitute in S

$44.1 = 0 + \frac{1}{2}×9.8×t^{2}$

=> $t^{2} = \frac{441}{49}$

=> t = 3

•°• Second stone will come down in 3-1 = 2 sec
(Given that second stone reaches downward 1 sec later that of first )

Now Sub t= 2 in same equation

$44.1 = 2u + \frac{1}{2}×9.8×4$
=> 44.1 = 2u + 19.6
=> 2u = 44.1 - 19.6
=> u = 12.25 m/s


•°• Initial Speed of 2nd stone = 12.25 m/sec

•°•°•°•°•°<><><<><>><><>°•°•°•°•°

Answer 2

Answer:

Explanation:

Hi there!

Here's the answer:

•°•°•°•°•°<><><<><>><><>°•°•°•°•°

We have,

Given,

S = 44.1 m

a = g = 9.8 m/s²

and for a freely falling body u =0

Substitute in S

=>  

=> t = 3

•°• Second stone will come down in 3-1 = 2 sec

(Given that second stone reaches downward 1 sec later that of first )

Now Sub t= 2 in same equation

=> 44.1 = 2u + 19.6

=> 2u = 44.1 - 19.6

=> u = 12.25 m/s

•°• Initial Speed of 2nd stone = 12.25 m/sec

•°•°•°•°•°<><><<><>><><>°•°•°•°•°