Question

a stone is dropped into water from bridge at the height 44.1 m above the sea. another stone is thrown vertically downward 1sec after both will strike the water simultaneously what would be the initial speed of the second stone?

Answer 1

Let ‘t’ be the time taken by both the stones to reach the water.

For the stone that is dropped, using,

S = ut + ½ at2

=> 441 = 0 + ½ (9.8)t2

=> t = 9.5 s

Let the other stone be thrown at initial speed ‘u’. It reaches the water in (9.5 – 1 =) 8.5 s.

Using,

S = ut + ½ at2

=> 441 = 8.5u + ½ (9.8)(8.52)

=> u = 10.2 m/s

So, the second stone is thrown at 10.2 m/s.

Answer 2

Answer:

Explanation:

x=44.1m

a=9.8m/s2

u=0

Wkt,

x=ut+1/2gt2

44.1=1/2*9.8*t2

t2=441/49

t=3 sec. So the second stone will come down in 3-1=2sec

Therefore, applying the same eqn again we get,

44.1=2u+1/2*9.8*4

44.1=2u+19.6

u=12.25m/s