Question

A ball is released from a height and it reaches the ground in 3s. if g=9.8m s-2 ,find:
(a) The height from which the ball was released,
(b) The velocity with which the ball will strike the ground.

Answer 1

Given, t = 3 sec

u = 0 m/s

a = g = 9.8m/s²

v = u + at

v = 0 + 9.8(3)

v = 29.4 m/s

v² - u² = 2as

(29.4)² - 0 = 2(9.8)s

s = (29.4)²/19.6

s = 44.1 m

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Answer 2

Answer:

height will be 44.1m whereas its final velocity would be 29.4m/s

Explanation:

GIVEN:

h=?

t=3s

initial velocity=0

final velocity=?

FORMULA:

h=v(initial)t +1/2gt (taking square of time)

final velocity=initial velocity +gt

SOLVE IT BY PUTTING VALUES IN FORMULA;-)

P.S:not sure about tge answer