Answer:
1.5 Amp is rated for 5 W so it would not be possible
Answer:

Explanation:
We have given that the battery has an internal emf of 9 volt
So E = 9 volt
Capacitance 
It is given that on the terminal voltage is only 80% of potential difference
So V = 0.8×9 = 7.2 volt
We know that energy stored in the capacitor is given by

x(2) = 30(2) = 60 m
y(2) = ½(-9.8)2² = -19.6 m
x(4) = 30(4) = 120 m
y(4) = ½(-9.8)4² = -78.4 m
x(6) = 30(6) = 180 m
y(6) = ½(-9.8)6² = -176.4 m
x(8) = 30(8) = 240 m
y(8) = ½(-9.8)8² = -313.6 m
x(10) = 30(10) = 300 m
y(10) = ½(-9.8)10² = -490 m