Question

a stone is dropped freely in river from a bridge. It takes 10s to touch the water surface in the river.( i ) the height of the bridge from the water level ( ii ) the distance covered by the stone in 4s ( g = 9.8 m /s² )​

Answer 1

QUESTION:-

a stone is dropped freely in river from a bridge. It takes 10s to touch the water surface in the river.( i ) the height of the bridge from the water level ( ii ) the distance covered by the stone in 4s ( g = 9.8 m /s² )​

EXPLANATION:-

• Initial velocity(u)= 0 m/s
• Acceleration=Acceleration due to gravity(a)=g = 9.8 m /s²
• Time(t)=10 s
• Height(h)=?
• Distance covered(s)= ?

(i) From equation of motion, h = ut + 1/2gt²

h = 0 × 5 + 1/2 × 9.8 × (10)²

 h = 1/2×9.8×100

 h = 490 m

∴ Height of bridge is 490 m

------------------------------

(ii)S = ut + 1/2gt²

 S = 0 + 1/2 × 9.8 × (4)²

 S = 1/2×9.8×16

S=78.4 m

∴ The distance covered by the stone is the first 4 s is 78.4  m.

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Answer 2

:ANSWER IS

IN CASE 1=122.5  M (METER)

IN CASE 2=44.1  M( METER)

:QUESTION IS

a stone is dropped freely in river from a bridge. It takes 10s to touch the water surface in the river.( i ) the height of the bridge from the water level ( ii ) the distance covered by the stone in 4s ( g = 9.8 m /s² )​

:STEP BY STEP ANSWER

=Given :  

Initial speed=u= 0 m/s

time=t=5sec

Case 1)

distance travelled = height of bridge=h m

From second equation of motion:

s=ut+1/2at²h

0xt +1/2 9.8 x 5 x5h

22.5m

∴The height of the bridge from water level  is 122.5m

----------------------------------------------------------------------

CASE 2

Distance travelled in 4sec

(1/2x9.8x4x4)

16x4.9

78.4 m

Distance travelled in last second :

Distance travelled in 5sec - distance travelled in 4 sec=122.5 -78.4=44.1m

∴The distance covered by stone in the last second is 44.1m