Question

I have some doubts in the following questions:
Chapter : WORK , ENERGY AND POWER

1. A body of mass 2 kg is dropped from the top of a tower of height 100 m. If the acceleration due to gravity is 10 m/s2 calculate the K.E at the end of 5 seconds.
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Answer 1

Answer:

$v = u + at$

here

1. u=0
2. a=10
3. t=5

ie v= 0 +10(5)

=50m/s

$k.e = \frac{1}{2} m {v}^{2}$

here

1. m=2kg
2. v=50

ie. ke =

$\frac{1}{2} \times 2 \times {50}^{2}$

=

$2500$

Answer 2

[tex]

v = u + atv=u+at

here

u=0

a=10

t=5

ie v= 0 +10(5)

=50m/s

k.e = \frac{1}{2} m {v}^{2}k.e=21mv2

here

m=2kg

v=50

ie. ke 21×2×502

=2500