Question

a ball drops from height it takes 1sec to cross the last 55m before from which it was dropped find answer​{g=10}

Answer 1

Let height is h from which the ball was dropped. after covering (h - 55) m, velocity of ball is v.

Initial velocity of ball, u = 0

using formula,

v² = u² + 2as

or, v² = 0 + 2g(h - 55) = 20(h - 55)

or, v = √{20(h - 55)}

ball takes 1 sec to cross the last 55 m.

so, t = 1, s = 55m and u = √{20(h-55)}

now, using formula ,

S = ut + 1/2 gt²

or, 55 = √{20(h - 55)} × 1 + 1/2 × 10 × 1²

or, 50 = √{20(h - 55)}

or, 2500 = 20(h - 55)

or, 125 = h - 55

or, h = 125 + 55 = 180m

hence, height of ball is 180m