You throw a baseball (mass 0.145 kg) vertically upward. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected. At what height above your hand does the ball have (a) half as much upward velocity, (b) half as much kinetic energy as when it left your hand

Explanation:

mass of baseball, m = 0.145 kg

initial velocity, u = 12 m/s upward

(a) final velocity, v = u / 2 = 6 m/s

Let the height is h.

Use third equation of motion

v² = u² - 2gh

6 x 6 = 12 x 12 - 2 x 9.8 x h

36 - 144 = - 19.6 x h

h = 5.51 m

(b) initial kinetic energy, K = 0.5 x m x u² = 0.5 x 0.145 x 12 x 12 = 10.44 J

Final kinetic energy, K' = K/2

0.5 x m x v² = 10.44 /2

0.5 x 0.145 x v² = 5.22

v = 8.5 m/s

v² = u² - 2gh

8.5 x 8.5 = 12 x 12 - 2 x 9.8 x h

72.25 - 144 = - 19.6 x h

h = 3.66 m

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j