The diagram given below shows a ski jump. A skier weighing 60kgf stands at A at the top of ski jump. He moves from A and takes off for his jump at B.

(a) Calculate the change in the gravitational potential energy of the skier between A and B.

(b) If 75% of the energy in part (a) becomes the kinetic energy at B, calculate the speed at which the skier arrives at B.

(Take g = 10 ms^-2).

Given,

Mass = 60 kg

(a)

$\text {Loss in potential energy} = \text{mg (h1 – h2)} \\[0.5em] = 60 \times 10\times (75-15) \\[0.5em] = 60 \times10 \times 60 \\[0.5em] = 3.6 \times 10^4 J \\[0.5em]$

(b) When kinetic energy at B is 75% of (3.6 × 10⁴)

$\text{Kinetic energy at B} = \dfrac{75}{100} \times 3.6 \times 10^4 \\[0.5em] = 27000J \\[0.5em] = 2.7 \times 10^4 J \\[0.5em]$

Since,

Kinetic energy = $\dfrac{1}{2}$ mv²

Substituting the values in equation we get,

$27000 = \dfrac{1}{2} \times 60 \times v^2 \\[0.5em] \Rightarrow 27000 = 1 \times 30 \times v^2 \\[0.5em] \Rightarrow v^2 = \dfrac{27000}{30} \\[0.5em] \Rightarrow v^2 = 900 \\[0.5em] \Rightarrow v = \sqrt{900} \\[0.5em] \Rightarrow v = 30 \\[0.5em]$

∴ The speed at which the skier arrives at B = 30ms^-1