Question

help me with this fast​

Answer 1

Energy at A

= 1/2 m v^2 + mgh

= m( 10^2/2 + 20 g)

= 10( 50 + 200)

= 2500

Kinetic energy at B= 2500 as potential energy would be 0 at B

At C

KE + potential = 2500

potential = mgh = 10(10)(10) = 1000

KE = 2500 -1000 = 1500

At D

KE + PE = 2500

PE at D= mgh

h = 3

PE = 10(10)(3) = 300

KE = 2500 -300 = 2200

Now as it has KE so it will go beyond D

✌✌✌Dhruv✌✌✌✌

Answer 2

kinetic energy at B=1/2mv^2+mgh

=1/2×10×10×10+10×10×20

=500+2000

=2500J

k.e. at C=1/2 mv^2+mgh

1/2×10×10×10+10×10×10

500+1000

1500J