Abody is projected vertically upwards with a speed ‘u’. find the (a) time taken and (b) the distance travelled by the body in reaching the topmost point. take a= -g.

(a) t = u/g

(b) Distance travelled = (u^2/g) - (u^2/2g)

Explanation:

(a)When it reaches its topmost point...the final speed v = 0...;

Apply the formula v = u + at

; 0 = u + (-g)t

So hence the time taken to reach the top most point; t = u/g

(b)Apply the formula...s = ut + 1/2at^2

where s is the distance travelled

also substitute t with the answer in (a)

s = u(u/g) + 1/2(-g)(u/g)^2

s = (u^2/g) - (u^2/2g)

0.55x745.7j and a calculator

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