Question

A body falling from the rest describes 44.1m in the last second of its fall find the time of it's fall and the height from which it fell​

Answer 1

Answer:

4.5s

99.225m

Explanation:

u=0 a=9.8

last second tavel=v=44.1m

s=v²-u²/2a

44.1²/19.6

99.225m=height

s=ut+1/2×at²

u=0 a=9.8

99.225=1/2×9.8×t²

t²=20.25

t=4.5s

Answer 2

Answer:

the time of it's fall is $t=4.5s$

the height from which it fell​ is $= 122.5m$

Explanation:

In physics, motion is a function of a body's location or rotation over time. Translation is defined as movement along a line or a curve. Rotation is a motion that modifies a body's orientation. In all scenarios, the velocity and direction speed of every point on the body are equal (time rate of change of velocity). Motion that involves both movement and rotation is the most generic type.

$s=ut+\frac{1}{2} *at^{2}$

$u=0$

$a=9.8$

$99.225=\frac{1}{2} *9.8*t^{2}$

$t^{2} =20.25$

$t=4.5s$

Therefore, the time of it's fall is $t=4.5s$

Given,

$g=9.8m/s^2$

$h=44.1m$

$t=1sec$

Using the second equation of motion

$s = ut+\frac{1}{2}* gt^2$

$44.1=u(1)+\frac{1}{2} *9.8*(1)^{2}$

$44.1 = u+4.9$

$u=39.2m/s (final velocity)$

Using the third motion equation,

$2gH=v^{2} -u^{2} (initial velocity is zero)$

$H=\frac{39.2*39.2}{2*9.8}$

the tower's height,

$H_{t} =H+h$

    $=78.4+44.1$

    $= 122.5m$

Therefore, the height from which it fell​ is $= 122.5m$

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