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
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?
1 answer
0 answers