Question

A body of mass 5kg is dropped from a height of 1m. Its kinetic energy just before reaching the ground is

Answer 1

V² = (2×10 × 1 )= 20 m/s

K E = max energy = 1/2 × 5 × 20 = 50 J

Answer 2

$\huge{\boxed{\boxed{\sf{\red{Answer:}}}}}$

${\boxed{\boxed{\sf{Given:}}}}$

➜ A body of mass 5 kg is dropped from a height of 1m.

${\boxed{\boxed{\sf{Find:}}}}$

➜ Find what is kinetic energy just before reaching the ground.

${\boxed{\boxed{\sf{According \:to\: the\: question:}}}}$

➜ Mass = 5 kg

➜ Height = 1 m

${\boxed{\boxed{\sf{Know \:terms:}}}}$

➜ $\sf{Initial\:velocity = (u)}$

➜ $\sf{Final\:velocity = (v)}$

➜ $\sf{Acceleration = (a)}$

➜ $\sf{Joules = (j)}$

${\boxed{\boxed{\sf{According\: to\: equation \:of \:motion:}}}}$

➜$\sf{v^2 = u^2 + 2 gh}$

➜ $\sf{v^2 = 0^2+ 20 h}$

➜ $\sf{v^2 = 20 m/sec}$

➜ $\sf{v = 1/2 mv^2}$ => kinetic energy of the body with velocity.

➜ $\sf{Kinetic \:energy = 1/2 \times 5 \times 20}$

➜ $\sf{Kinetic \:energy = 50\: J}$

Therefore, 50 joules its kinetic energy just before reaching the ground.