Question

A stone of 1 kg dropped from height h and strikes ground with velocity of 25m/s.Find kinetic energy

Answer 1

here it is

for height

W = mg = (4)(9.8) = 39.2 N

here we use

2gh = vf^2 - vi^2

where vf and vi are the final and initial velocities

Vi =0 since when it starts falling its initial velocity will be zero

vf = 25 m/s since velocity when it strikes ground is given

now substituting in equation

2(9.8)(h) = (25)^2 - (0)^2

2(9.8)h = 625

h = 625/(2*9.8)

h = 31.88 = 32 m ANSWER

Answer 2

Answer:

312.5 Joules

Explanation:

Kinetic energy= 1/2mv square

m= 1kg

V= 25m/s

Therefore,

KE= 1/2 x 1 x25 x 25

= 625/ 2

=312.5 Joules