Question

On a cold day, hailstones fall with a velocity of 2i − 6k m s−1 . If a cyclist travels through the hail at 10i m s−1 , what is the velocity of the hail relative to the cyclist? At what angle are the hailstones falling relative to the cyclist?

Answer 1

Answer:

The velocity of the hail relative to the cyclist is $V_{hc}=-8\hat{i}-6\hat{k}$

The angle at which hailstones falling relative to the cyclist is $\theta = 36.86^\circ$

Step-by-step explanation:

Given : On a cold day, hailstones fall with a velocity of $2\hat{i}-6\hat{k}\text{ m/s}$ . If a cyclist travels through the hail at $10\hat{i} \text{ m/s}$.

To find : What is the velocity of the hail relative to the cyclist and At what angle are the hailstones falling relative to the cyclist?

Solution :

The velocity of the hailstone falls is $V_h=2\hat{i}-6\hat{k}\text{ m/s}$

The velocity of the cyclist travels through the hail is $V_c=10\hat{i} \text{ m/s}$

The velocity of the hail relative to the cyclist is given by,

$V_{hc}=V_h-V_c$

Substitute the value in the formula,

$V_{hc}=2\hat{i}-6\hat{k}-10\hat{i}$

$V_{hc}=-8\hat{i}-6\hat{k}$

So, The velocity of the hail relative to the cyclist is $V_{hc}=-8\hat{i}-6\hat{k}$

Now, The angle of hails falling relative to the cyclist is given by

$\theta = \tan^{-1}(\frac{-6}{-8})$

$\theta = \tan^{-1}(\frac{3}{4})$

$\theta = 36.86^\circ$

So, The angle at which hailstones falling relative to the cyclist is  $\theta = 36.86^\circ$