posted by Ray on .
Suppose you throw a 0.058-kg ball with a speed of 11.0 m/s and at an angle of 31.5° above the horizontal from a building 12.3 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?
the throwing angle has no effect
the KE at impact is the initial KE plus the KE gained from the gravitational potential (m g h)
KE = (.5 * .058 * 11.0^2) + (.058 * 9.8 * 12.3)
v^2 = (2 * KE) / .058