Question

A lead ball is dropped 30 feet from a church tower. If the acceleration due to gravity is 9.8m s^-2, how long does it take the ball to fall to the ground?

Answer 1

final velocity of a falling object = 0

and acceleration of a falling object is negative

now using the expression ( Newton's 3rd equation of motion)

${v}^{2} = {u}^{2} - 2gs$

where the parameters have their usual meaning

making

${u}^{2} \: the \: subject$

thus

${u}^{2} = 2gs$

$u = \sqrt{2gs}$

from the question,

g= 9.8m/s^2

s = 30m

substitute them into the equation

$u = \sqrt{2(9.8 \times 30)}$

$u = \sqrt{588}$

u = 24.25m/s

but the question is demanding for time but not the initial velocity

so substitute them in the

Newton's first equation of motion

where

V = u - gt

but from the question,

v= 0 and the acceleration is negative because it's a free fall

substitute the values into the equation

0 = 24.24 -9.8t

making t the subject

9.8t = 22.24

dividing through by 9.8

9.8t/9.8 = 22.24/9.8

t = 2.3s . therefore, it would take 2.3s for the ball to hit the ground.