Answer
given,
range of the projectile = 4.3 m
time of flight = T = 0.829 s
v = 5.19 m/s
vertical component of velocity of projectile
v_y = gt'
a) Launch angle
θ = 38°
b) initial speed of projectile
v = 6.59 m/s
c) maximum height reached by the projectile
Answer:
Part a)
Part b)
Part d)
As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different
On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.
Explanation:
Part a)
By Guass law we know that
Part b)
Outside the outer cylinder we will again use Guass law
Part d)
As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different
On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.
