A ball is thrown horizontally from the top of a 20-m high hill. It strikes the ground at an angle of 45o45o . With what speed was it thrown?

V = 19.79 m/s

Explanation:

given,

height of the hill, h = 20 m

angle of strike,θ = 45°

Speed of thrown = ?

using equation of motion for time calculation

  t = 2.02 s

Horizontal velocity = v_x = V

Vertical velocity = v_y= g t

                              v_y = 9.8 x 2.02 = 19.79 m/s

now, we know,

  V = 19.79 m/s

Hence, the speed at which ball was thrown is equal to 19.79 m/s.

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