Question

A stone is dropped freely in a river from a bridge. It takes 5s to touch the water surface in the

river. Calculate :

1. the height of the bridge from the water level,

2. the distance covered by the stone in the first 2s (g = 9.8 msˉ²)​

Answer 1

   Given :  

Initial speed=u= 0 m/s

time=t=5sec

Case i)

distance travelled = height of bridge=h m

From second equation of motion:

s=ut+1/2at²h

=0xt +1/2 9.8 x 5 x5h

=122.5m

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

Case ii)

Distance travelled in first 2sec

= (1/2x9.8x2×2)

=4x4.9

=19.6 m.

Answer 2

Answer:

1] 122.5m

2] 44.1m

⭐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+1/2 at^2h

s=0xt +1/2 9.8 x 5 x 5 h  

s=122.5m  

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