I throw a ball straight up with an initial velocity of 2m/s. How high is it after a fifth of a second?

assign each variable to a value in s = ut + 1/2at2u = 2m/s t=0.2sa=-9.81m/s2so s = (2 x 0.2) + 1/2(-9.81)(0.2)2so s =

HA
Answered by Henry A. Physics tutor

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