Question

Q2. A stone of mass 20g is dropped 1p
from a cliff. What will be its KE when
it has fallen 100 m? *​

Answer 1

Answer:

Given:-

•Distance(s)=100m

•Acceleration due to gravity(g)=9.8m/s²

•Initial velocity(u)=0

To find:-

•Speed of the stone when it has fallen 100m

Solution:-

By using the 3rd equation of motion,when u=0,we get:-

=>v²=2gs

=>v²=2×9.8×100

=>v²=1960

= > v = \sqrt{1960}=>v=

1960

=>v=44.27m/s(approx)

Thus,speed of the stone when it has fallen 100m is 44.27m/s square.

Answer 2

Answer:

when objects are dropping there initial velocity will be equals to zero

let's find the final velocity

${v}^{2} = {u}^{2} + 2as \\ v = \sqrt{2 \times 10 \times 100} \\ v = \sqrt{2000} \\ v = 44.7m {s}^{ - 1}$

kinetic energy

$\frac{1}{2}m {v}^{2} \\ \frac{1}{2} \times 0.02 \times 2000 \\ 20j$