Question

a ball is thrown up at a speed 4 m/s. find maximum height reached by the ball take g 10m/s

Answer 1

U = 4 m/s V = 0   a = 10 m/ s^2. S = ?        ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~     v ^2 = u ^2 + 2as                                                   0 = 16 m/s + 2 × 10 m/s^2 × s                    S  = - 16 / 20.  M                                          S  =  0.8   M .

Answer 2

We know initial and final velocity

So, by applying eq. of motion

V^2 - u^2 = 2as

We will get distance (s) to be 0.8 m

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