JEE Main 18 March 2021 Shift 1 Solved Paper

Given,
The mass of the particle, m=10kg
The speed of the particle at point A,vA=0m/s
The elevation of the point A from the point B,hA=5+hB
Let's consider the speed of the particle at point B=vBUsing the law of conservation of energy.
Energy at point A= Energy at point B
‌

1

2

mvA2+mghA=‌

1

2

mvB2+mghB
Substituting the values in the above equations, we get
vB=10m∕s=xm∕s
∴‌‌x=10