Question

plss solve this question .....it is from today's physics paper( 10ICSE boards)

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Answer 1

(i)
Potential energy = mgh
= (10)(10)(3) = 300 Joule
Kinetic energy = mv²/2
v² - u² = 2gh
v² - 0² = 2 (10) 3
v² = 60
KE = mv²/2 = 10(60)/2
KE = 300 joule
Total Energy = 600 Joule
where m is the mass is the body
g is the acceleration due to gravity
h if the height.

(ii)
Just before hitting the ground.

So, Total Energy just before hitting the ground = KE + PE
v² - u² = 2gh
v² - 0² = 2 (10) 5
v² = 100
KE = mv²/2 = 10(100)/2
KE = 500 Joule
PE = mgh = 10 (10) 5
= 500 Joule
Total Energy = 1000 Joules

Thankyou!!!

Answer 2

Since body is allowed to freely fall, it gains velocity

v² - u² = 2gs

v² = 2(10)(3)

v² = 60 m²/s²

KE at a height 2m

KE = 1/2 m v²

KE = 10 X 60 / 2

KE = 300 J

PE at height 2m

PE = m g h'

PE = 10(10)(2)

PE = 200 J

Total Mechanical energy at height 2m

E = KE + PE = 500 J

(ii) Velocity before hitting the ground

v² = 2 g h

v² = 2(10)5

v² = 100 m²/s²

KE = 1/2 m v²

KE = 10 X 100/2

KE = 500 J