Question
the speed of a bowling ball is 2.0m/s as it falls past a window of a tall building. using conservation of energy ,determine how fast it was going as it past another window 3.0m below.
Answers
Steve
mgh + 1/2 mv^2 stays constant, so
mgh1 + 1/2 m*2^2 = mg(h1-3) + 1/2 mv2^2
divide out all the m's and you have
9.8*h1 + 2 = 9.8(h1-3) + 1/2 v2^2
2 = -9.8*3 + 1/2 v2^2
v2^2 = 62.8
v2 = 7.92
mgh1 + 1/2 m*2^2 = mg(h1-3) + 1/2 mv2^2
divide out all the m's and you have
9.8*h1 + 2 = 9.8(h1-3) + 1/2 v2^2
2 = -9.8*3 + 1/2 v2^2
v2^2 = 62.8
v2 = 7.92