Question

7 A ball has a mass of 0.5 kg. Dropped from a cliff top, the all strikes the sea below at a velocity of 10 m/s. /s. a) What is the kinetic energy of the ball as it is about to b) What was its potential energy before it was c) From what height was the ball dropped? strike the sea? dropped? Repeat stages a, b, and c above to find the height t is a d) A stone of mass 1 kg also strikes the sea at 10 m/s. ch from which the stone was dropped. Figure 7 3,

Answer 1

K E = 1/2 mv^2 = 1/2 times 0.5 times (10 m/s) = 1/2 times 1/2 times 100 joule = 25 joule (b) P E = m g h = 1 kg/2 times 9.8 m/s^2 times h fro newton eqn of motion v^2 = u^2 - 2gh asymptoticallyequalto u = 0, 100 m^2/s^2 = - 2 times 9.8m/s^2 times h h = 5.10 m P E = 4.9 times 5.10 = 24.99 joule (c) v^2 = u^2 - 2as rightdoublearrow 100 m^2/s^2 = - 2 times 9.8 m/s^2 times h h = 5.10 m for 1 kg ball (a) K E = 1/2 times 100 joule = 50 joule (b) p E = 1 times 9.8 times 5.10 = 49.98 joule (c) h = 5.10 m