Question

Q.6
A body of mass 5Kg is dropped from the top of a tower of height 100m. What will be kinetic energy at the end of 2 second. (Given g =9.8m/s2)
​

Answer 1

Answer:

velocity does not change in freefall

because direction remains same

so initial velocity is equal to final velocity

we are taking g=10m/s2

it does not change anser much it give approx value to real value

Explanation:

mass=5kg

v=not given

kinetic energy =?

$s = ut + \frac{1}{2} a {t}^{2} \\ 100 = u \times 2 + \frac{1}{2}10 \times 4 \\ \frac{100}{20} = u2 \\ \frac{5}{2} = u \\ 2.5= u$

mass=5kg

v=2.5m/s

kinetic energy =?

$kinetic \: energy = \frac{1}{2}m {v}^{2} \\ = \frac{1}{2} \times 5 \times 2.5 \\ = \frac{12.5}{2} \\ = 6.25$

here ur answer