Question #22109

During the hammer throw at a track meet, an 8.0 kg hammer is accidentally thrown straight up. If 784.0 J of work were done on the hammer to give it vertical velocity, how high will it rise?

Expert's answer

The energy conservation law:

T+U = const

T - kinetic energy

U - potential energy

T = m*v^2/2

U = m*g*h

After the throw:

A = T

A + 0 = U + 0

A = m*g*h

h= A/(m*g) = 784/(8*9.8)= 10 m

T+U = const

T - kinetic energy

U - potential energy

T = m*v^2/2

U = m*g*h

After the throw:

A = T

A + 0 = U + 0

A = m*g*h

h= A/(m*g) = 784/(8*9.8)= 10 m

## Comments

## Leave a comment