Question

A stone is dropped freely in the 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 stone in the last second

Answer 1

$\huge{\underline{\tt{\blue{Answer}}}}$

1] 122.5m

2] 44.1m

$\huge{\underline{\tt{\orange{Solution}}}}$

⭐Given⭐

Initial speed=u= 0 m/s

time=t=5sec

⭐Case 1⭐

Distance travelled = height of bridge= h m

From second equation of motion:

$\boxed{s=ut+\dfrac{1}{2}at^2h}$

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

Answer 2

Answer:

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.