Question

a body of mass 50 kg is dropped from a ht. of 20m.find K.E & P.E when t = 1s

Answer 1

at t = 1 s
height fallen
h = 0.5×g×t^2 = 0.5× 9.8×1^2 = 4.9 m
so height from ground H = 50 m -4.9 m = 45.1 m
velocity after 1s
$v = 9.8 \times 1s = 9.8m {s}^{ - 1}$
so potential energy
$u \: = mgh = 50 \times 9.8 \times 45.1 = 22099j$
= 22099 J
now kinetic energy after 1s
KE = 0.5mv^2
= 0.5 ×50×(9.8)^2
= 2401 J