Physics
posted by James on .
Suppose you throw a 0.054kg 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?

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