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 speed water simultaneously. What was the initial speed of the second stone?(a) 12.25/s(b) 14.75m/s(c) 16.23m/s(d) 17.15m/s
Hrishant Goswami , 11 Years ago
Grade 10
2 Answers
Keerthi Raj
Last Activity: 11 Years ago
placing origin of the coordinate system at the point of dropping.
-44.1=-4.9t2 ==>t=3s
-44.1=2u-19.6 ==> u=-12.25m/s
Yash Deshpande
Last Activity: 6 Years ago
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
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