A stone is dropped into water from a bridge 44.1 m above the water. Another stone is thrown vertically downward 1 sec later. Both strike the water simultaneously. What was the initial speed of the second stone ?

(1) 12.25 m/s

(2) 14.75 m/s

(3) 16.23 m/s

(4) 17.15 m/s

Concept Videos :-

#26-Equations-for-Uniform-Acceleration
#27-Equations for Uniform Acceleration 28Calculus Method29
#28-Sign Convention
#29-Solved Examples 16
#31-Solved-Examples-18
#32-Solved Examples 19
#33-Solved Examples 20
#34-Solved-Examples-21
#35 Solved Examples 22
#30-Solved Examples 17

Concept Questions :-

Uniformly accelerated motion

(1) Time taken by first stone to reach the water surface from the bridge be t, then

h=ut+12gt244.1=0×t+12×9.8t2

t=2×44.19.8=3sec

Second stone is thrown 1 sec later and both strikes simultaneously. This means that the time left for second stone =31=2sec

Hence 44.1=u×2+129.8(2)2

44.119.6=2uu=12.25m/s  

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