Question

An ice skier is sliding from a point $\text{A}$ as shown in the figure below. If the slider starts from rest and acceleration due to gravity is $10\text{m}/s^2$, then what is the velocity of the skier at point $\text{B}$.

• A. $10\sqrt{5}\text{m/s}$
• B. $10\sqrt{2}\text{m/s}$
• C. $15\sqrt{5}\text{m/s}$
• D. $20\sqrt{2}\text{m/s}$

Answer 1

The correct option is D $20\sqrt{2}\text{m/s}$

On applying the principle of energy conservation, loss in potential energy of skier equals gain in the kinetic energy of the skier.
$h_A=$ Height of $\text{A}$ from the ground =$50\text{m}$
$h_B=$ Height of $\text{B}$ from the ground=$10\text{m}$

Thus,
$(mg{h}_A-mg{h}_B)=(\frac{1}{2}m{V}_B^2-\frac{1}{2}m{V}_A^2)$
As the initial velocity at point A is zero
$mg(h_A-h_B)=\frac{1}{2}m{V}_B^2$
$⟹g(50-10)=\frac{1}{2}{V}_B^2$
$⟹\frac{1}{2}{V}_B^2=\left(10\right)\times \left(40\right)$
$⟹V_B^2=2\times 400=800$
$⟹V_B=\sqrt{800}=20\sqrt{2}\text{m/s}$
Hence, the speed of the skier at point $\text{B}$ is $20\sqrt{2}\text{m/s}$