Mass = 60 kg
(a) Loss in potential energy = mg(h1 – h2)
= 60 × 10 × (75 – 15)
= 60 × 10 × 60
= 36000
= 3.6 × 104 J
(b) When kinetic energy at B is 75% of (3.6 × 104)
KE at B = 75\over 100 × 3.6 × 104
= 27000 J
= 2.7 × 104 J
Since, KE = 1\over 2×m×v2
⇒ 27000 = 1\over 2×60×v2
⇒ v2 = 27000\over 30
⇒ v2 = 900
⇒ v = √900 = 30 ms-1
∴ The speed at which the skier arrives at B = 30 ms-1