You are at the top of a 500 meter tower and drop a 5kg hammer. Calculate the potential and kinetic energies and total energy at the end of each second of free fall and at the moment of impact.

t D=5t2 V=at PE=mgh KE=1/2mv2
Elapse Distance Downward Potential Kinetic
time Fallen Velocity Energy Energy
0 sec meter/sec m /
1 s e c
2 s e c
3 s e c
4 s e c
5 s e c
6 s e c
7 s e c

You are at the top of a 500 meter tower and drop a 5kg hammer. Calculate the potential and kinetic energies and total energy at the end of each second of free fall and at the moment of impact.
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t D=5t2 V=at PE=mgh KE=1/2mv2
Elapse Distance Downward Potential Kinetic
time Fallen Velocity Energy Energy
0 sec meter/sec m /
1 s e c
2 s e c
3 s e c
4 s e c
5 s e c
6 s e c
7 s e c
50 0 m

The acceleration of the hammer is g = 9.8 m/s2

Where s is the distance fallen. Then the height of the hammer is

The potential energy is mgh, and the kinetic energy is
The total energy is PE+KE.

T=0, v= 9.8 0=0, h=500m, PE= 5 9.8 500= 24500 J, KE=0, TE=PE+KE= 24500J.

T=1, v = 9.8 1=9.8 m/s, s= m, h= 500- 4.9 = 495.1 m,
PE= 5 9.8 495.1 = 24259.9 J. KE= J, TE = 24500J.

T=2, v = 9.8 2=19.6 m/s, s= m, h= 500- 19.6 = 480.4 m,
PE= 5 9.8 480.4 = 23539.6 J. KE= J, TE = 24500J.

T=3, v = 9.8 3=29.4 m/s, s= m, h= 500- 44.1 = 455.9 m,
PE= 5 9.8 455.9 = 22339.1 J. KE= J, TE = 24500J.

T=4, v = 9.8 4=39.2 m/s, s= m, h= 500- 78.4 = 421.6 m,
PE= 5 9.8 421.6 = 20658.4 J, KE= J, TE = 24500J.

T=5s, v = 9.8 5=49 m/s, s= m, h= 500- 122.5 = 377.5 m,
PE= 5 9.8 377.5 = 18497.5 J, KE= J, TE = 24500J.

T=6s, v = 9.8 6=58.8 m/s, s= m, h= 500- 176.4= 323.6 m,
PE= 5 9.8 323.6 = 15856.4 J. KE= J, TE = 24500J.

T=7s, v = 9.8 7=69.6 m/s, s= m, h= 500- 240.1 = 259.9 m,
PE= 5 9.8 259.9 = 12735.1 J. KE= J, PE = 24500J.

Please see the attached file for the complete solution