The time to drop from 1.3m and velocity after height at 0.7 mis mathematically given as
t1 = 0.515 sec
v0 = 3.7m/s
Question Parameters:
A 0.045-kg golf ball is dropped from rest.
Is dropped from a height of 1.3m and comes back up at a height of .7m
Generally, the equation for the time to drop from 1.3m is mathematically given as
yf - yi = vi t + a t^2/ 2
Therefore
0 - 1.3 = 0 - 9.8 t^2 /2
t1 = 0.515 sec
Hence, after its max height is 0.7 m,
vf^2 - vi^2 = 2 a (yf - yi)
0^2 - (v0)^2 = 2(-9.8)(0.7 - 0)
v0 = 3.7m/s
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