Question

Establish the equation for position time relation (i.e., s=ut + (1/2)at^2) using velocity time graph. How is the
equation modified when
(a) a body is just dropped from some height, (b) a body is thrown vertically upwards with some velocity?​

Answer 1

Answer:

here is the answer

Explanation:

a) s=0+1/2(gt^2)

s= 1\2[gt^2]

b)if a body thrown up with a velocity then it comes down tho ithe point with upwards velipotha

s=ut+1/2(gt^2)

hope you understand mam

Answer 2

Answer:

hope it help you

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