A stone is thrown vertically upward with a speed of 28m/s. (a) find the maximum height reached by the?

given u= 28./s and v=0 ( as at maximum height the stone will come to rest instantenously)

using, v²= u²+2as { v and u as final and initial velocity, a- accleration (here gravity due to accleration) and s- max. height}

replacing a by -g as the stone is moving up against accleration due to gravtiy

and s is the max. height

so substituting the values,

(0)²= (28)²+2*9.8*s

so

784/19.6= s

maximum height = 40metres.

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