Question

Pls Pls Pls answer this question I will mark as brainliest


A stone is thrown vertically upwards with an initial velocity of 30m/S and acceleration of 10m/S^2 . Find the maximum height reached by the stone. What is the net displacement and total distance covered by the stone when it reaches back the ground

Answer 1

initial velocity, u=30m/s
acceleration,  g=10m/s²
the final velocity, v=u+at
at maximum height, v=0
that is, u+at=0
so,  30m/s=10 x t
∴ time, t= 3s
by displacement formula, s=ut-1/2gt²
∴  s=30x3-1/2(10)3²
    s=90-45
    so, the maximum height=45m//
since it falls down and comes back to the same position, net displacement is zero.//
the total distance covered by the stone=45+45
                                                             =90m.//
hope it helps....