Question

A 10 kg body falls from rest, with negligible interaction with its surrounding (no friction). determine its velocity after it falls 5 m

Answer 1

The body falls under the acceleration due to gravity , which is constant 9.8 or 10
For constant acceleration
V^2 - u^2 = 2as
U=0 (the body is dropped from rest)
a=g= +10 (along the acceleration due to gravity)
S = 5
Now by substituting the values
V^2 = 2(10)(5)
V^2 = 100
√V^2 = √100
V = 10
.Thus the velocity of the body is 10m/s

Answer 2

Δh = 1/2*V*t
since V = g*t, then :
Δh = 1/2*V*t = 1/2g*t*t = 1/2g*t^2
2*5 = 9.8*t^2
t = √10/9.8 = 1.01 sec
V= g*t = 9.8*1.01 = 9.90 m/sec

or by applying energy conservation principle :
m*g*h = 1/2m*V^2
m cross
V = √2gh = √98 = 9.90 m/sec
Hope it helps
Mark as brainlist!!!!❤️❤️