Asked by James
Suppose you throw a 0.054-kg ball with a speed of 11.5 m/s and at an angle of 31.8° above the horizontal from a building 13.1 m high.
(a) What will be its kinetic energy when it hits the ground?
(b) What will be its speed when it hits the ground?
(a) What will be its kinetic energy when it hits the ground?
(b) What will be its speed when it hits the ground?
Answers
Answered by
Delbarre
conservation of energy
mgh + (1/2)m(Vi)² = (1/2)m(Vf)²
2(gh + (1/2)(Vi)²) = (Vf)²
square root( 2gh +(Vi)²) = (Vf)
square root( 256.76 + 132,25) = (Vf)
square root( 389,01) = (Vf)
square root( 389,01) = (Vf)
19.7m/s = Vf
your kinetic energy is
(0.5)(0.054)(19.7)² = 10.5 J
enjoy the answer
mgh + (1/2)m(Vi)² = (1/2)m(Vf)²
2(gh + (1/2)(Vi)²) = (Vf)²
square root( 2gh +(Vi)²) = (Vf)
square root( 256.76 + 132,25) = (Vf)
square root( 389,01) = (Vf)
square root( 389,01) = (Vf)
19.7m/s = Vf
your kinetic energy is
(0.5)(0.054)(19.7)² = 10.5 J
enjoy the answer