Solved 2019 Question Paper ICSE Class 10 Physics — Question 1

(i) The speed of the body when at a height of 2m above the ground i.e., after falling through 3 m is :

V² = U² + 2gh = (0) + (2 x 10 x 3)
= 60

∴ V = √(60) m s^-1

Mechanical energy at 2m height = P.E. + K.E. = mgh + $\dfrac{1}{2}$ mv²
= (10 x 10 x 2) + ($\dfrac{1}{2}$ x 10 x 60)
= 200 + 300
= 500 J

Hence, total mechanical energy possessed by the body at the height of 2 m = 500 J

(ii) The kinetic energy on reaching the ground will be same as the potential energy at it's highest point from which it started falling.

Mechanical energy = Initial potential energy

= m x g x h = 10 x 10 x 5 = 500 J

Hence, kinetic energy possessed by the body just before hitting the ground = 500 J