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Free fall with initial velocity


From page 15 we learn furthermore

y(t) = v0yt - gt2/2

Hence, at the instant τ we have in the case of the golf ball

h = y(τ) = v0τ - gτ2/2 = v0(v0/g) - g(v0/g)2/2 = v02/2g

which leaves us with

v02 = 2gh

With h=3.07±0.15 m and g=9.8 m/s2, Walter Lewin arrives for the initial value of the velocity at

v02 = 2 ( 9.8 m/s2 ) × ( 3.07±0.15 m ) = 60.2±2.9 m2/s2

He does not calculate v0, because the future formulas only involve the square of v0.






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